Logarithms Questions in English

(D) The logarithmic function $y = \log_a x$ is defined under the following conditions:
$1$. The base $a$ must be a positive real number,i.e.,$a > 0$.
$2$. The base $a$ cannot be equal to $1$,i.e.,$a \neq 1$.
$3$. The argument $x$ must be positive,i.e.,$x > 0$.
Therefore,for the expression to be defined,$a$ must be any positive real number except $1$.

(C) Suppose,if possible,$\log_{2} 7$ is rational,say $p/q$ where $p$ and $q$ are integers,prime to each other.
Then,$\frac{p}{q} = \log_{2} 7 \implies 7 = 2^{p/q} \implies 2^{p} = 7^{q}$.
This is a contradiction because the $L.H.S$ is an even number (power of $2$) and the $R.H.S$ is an odd number (power of $7$).
Since it cannot be expressed as a ratio of two integers,$\log_{2} 7$ is an irrational number.
Obviously,$\log_{2} 7$ is not an integer and hence not a prime number.

(B) We know that for a fixed number $\alpha > 1$,the value of $\log_{b}\alpha = \frac{\ln \alpha}{\ln b}$.
Since $\ln \alpha$ is a positive constant,the value of $\log_{b}\alpha$ is inversely proportional to $\ln b$.
Comparing the bases: $10 > 3 > e \approx 2.718 > 2$.
Therefore,$\ln 10 > \ln 3 > \ln e > \ln 2$.
Taking the reciprocal,we get $\frac{1}{\ln 10} < \frac{1}{\ln 3} < \frac{1}{\ln e} < \frac{1}{\ln 2}$.
Multiplying by $\ln \alpha$,we get $\log_{10}\alpha < \log_{3}\alpha < \log_{e}\alpha < \log_{2}\alpha$.
Thus,the correct increasing order is $\log_{10}\alpha, \log_{3}\alpha, \log_{e}\alpha, \log_{2}\alpha$.

(B) Using the change of base formula $\log _{a}b = \frac{\log b}{\log a}$:
$\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{7}8 \cdot \log _{8}9$
$= \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} \cdot \frac{\log 9}{\log 8}$
$= \frac{\log 9}{\log 3}$
$= \log _{3}9 = \log _{3}(3^{2}) = 2 \cdot \log _{3}3 = 2 \cdot 1 = 2$.

(C) Given expression: $\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}}$
First,simplify the inner radical: $\sqrt{7 \sqrt{7 \sqrt{7}}} = (7 \cdot (7 \cdot 7^{1/2})^{1/2})^{1/2} = (7 \cdot (7^{3/2})^{1/2})^{1/2} = (7 \cdot 7^{3/4})^{1/2} = (7^{7/4})^{1/2} = 7^{7/8}$.
Now,substitute this back into the expression: $\log_{7} \log_{7} (7^{7/8})$.
Using the property $\log_{b} (b^x) = x$,we get $\log_{7} (7/8)$.
Using the property $\log_{b} (m/n) = \log_{b} m - \log_{b} n$,we get $\log_{7} 7 - \log_{7} 8$.
Since $\log_{7} 7 = 1$ and $8 = 2^3$,the expression becomes $1 - \log_{7} (2^3)$.
Using the power rule $\log_{b} (a^n) = n \log_{b} a$,we get $1 - 3 \log_{7} 2$.

(C) Let the expression be $E = 7 \log \left( \frac{16}{15} \right) + 5 \log \left( \frac{25}{24} \right) + 3 \log \left( \frac{81}{80} \right)$.
Using the property $n \log a = \log a^n$,we get:
$E = \log \left( \left( \frac{16}{15} \right)^7 \times \left( \frac{25}{24} \right)^5 \times \left( \frac{81}{80} \right)^3 \right)$.
Expressing the numbers in terms of prime factors:
$16 = 2^4, 15 = 3 \times 5, 25 = 5^2, 24 = 2^3 \times 3, 81 = 3^4, 80 = 2^4 \times 5$.
Substituting these values:
$E = \log \left( \frac{(2^4)^7}{(3 \times 5)^7} \times \frac{(5^2)^5}{(2^3 \times 3)^5} \times \frac{(3^4)^3}{(2^4 \times 5)^3} \right)$.
$E = \log \left( \frac{2^{28}}{3^7 \times 5^7} \times \frac{5^{10}}{2^{15} \times 3^5} \times \frac{3^{12}}{2^{12} \times 5^3} \right)$.
Grouping the powers of $2, 3,$ and $5$:
$E = \log \left( \frac{2^{28}}{2^{15} \times 2^{12}} \times \frac{3^{12}}{3^7 \times 3^5} \times \frac{5^{10}}{5^7 \times 5^3} \right)$.
$E = \log \left( \frac{2^{28}}{2^{27}} \times \frac{3^{12}}{3^{12}} \times \frac{5^{10}}{5^{10}} \right) = \log (2^1 \times 1 \times 1) = \log 2$.

(D) Given: $\log_{4}5 = a$ and $\log_{5}6 = b$.
Using the change of base formula,$ab = \log_{4}5 \times \log_{5}6 = \log_{4}6$.
Since $\log_{4}6 = \frac{\log_{2}6}{\log_{2}4} = \frac{\log_{2}(2 \times 3)}{2} = \frac{1 + \log_{2}3}{2}$.
So,$ab = \frac{1 + \log_{2}3}{2}$.
Multiplying by $2$,we get $2ab = 1 + \log_{2}3$,which implies $\log_{2}3 = 2ab - 1$.
Therefore,$\log_{3}2 = \frac{1}{\log_{2}3} = \frac{1}{2ab - 1}$.

(D) Given the equation: $\log _k x \cdot \log _5 k = \log _x 5$
Using the change of base formula $\log _a b \cdot \log _b c = \log _a c$,we get:
$\log _5 x = \log _x 5$
Let $y = \log _x 5$. Then $\log _5 x = 1/y$.
So,$1/y = y$,which implies $y^2 = 1$.
Therefore,$y = 1$ or $y = -1$.
If $\log _x 5 = 1$,then $x^1 = 5$,so $x = 5$.
If $\log _x 5 = -1$,then $x^{-1} = 5$,so $x = 1/5$.
Thus,$x$ can be $5$ or $1/5$.

(C) Given the equation: $\log_{5} a \cdot \log_{a} x = 2$
Using the change of base formula $\log_{b} a = \frac{\log_{k} a}{\log_{k} b}$,we can write:
$\left( \frac{\log x}{\log a} \right) \cdot \left( \frac{\log a}{\log 5} \right) = 2$
$\frac{\log x}{\log 5} = 2$
Applying the base change property in reverse:
$\log_{5} x = 2$
Converting to exponential form:
$x = 5^{2} = 25$

(C) Given $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$
First,simplify $\log _4 256 = \log _4 (4^4) = 4 \log _4 4 = 4 \times 1 = 4$.
Next,simplify $2 \log _{\sqrt{2}} 2 = 2 \log _{2^{1/2}} 2 = 2 \times \frac{1}{1/2} \log _2 2 = 2 \times 2 \times 1 = 4$.
Substitute these values back into the expression for $A$:
$A = \log _2 \log _2 (4) + 4$
Since $\log _2 4 = \log _2 (2^2) = 2$,we have:
$A = \log _2 (2) + 4$
Since $\log _2 2 = 1$,we get:
$A = 1 + 4 = 5$.

(D) Given that ${\log _{10}}x = y$.
We need to evaluate ${\log _{1000}}{x^2}$.
Using the change of base formula and properties of logarithms:
${\log _{1000}}{x^2} = \frac{{\log _{10}}{x^2}}{{\log _{10}}{1000}}$
$= \frac{2{\log _{10}}x}{{\log _{10}}{{10}^3}}$
$= \frac{2{\log _{10}}x}{3{\log _{10}}10}$
Since ${\log _{10}}10 = 1$,we have:
$= \frac{2}{3}{\log _{10}}x = \frac{2}{3}y$.

(C) Given $a = \log_{24} 12 = \frac{\log 12}{\log 24} = \frac{2\log 2 + \log 3}{3\log 2 + \log 3}$.
$b = \log_{36} 24 = \frac{\log 24}{\log 36} = \frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}$.
$c = \log_{48} 36 = \frac{\log 36}{\log 48} = \frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}$.
Multiplying these,$abc = \left(\frac{2\log 2 + \log 3}{3\log 2 + \log 3}\right) \times \left(\frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}\right) \times \left(\frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}\right)$.
Canceling common terms,$abc = \frac{2\log 2 + \log 3}{4\log 2 + \log 3}$.
Then $1 + abc = 1 + \frac{2\log 2 + \log 3}{4\log 2 + \log 3} = \frac{4\log 2 + \log 3 + 2\log 2 + \log 3}{4\log 2 + \log 3} = \frac{6\log 2 + 2\log 3}{4\log 2 + \log 3} = 2 \times \frac{3\log 2 + \log 3}{4\log 2 + \log 3}$.
Since $b = \frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}$ and $c = \frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}$,we have $bc = \frac{3\log 2 + \log 3}{4\log 2 + \log 3}$.
Thus,$1 + abc = 2bc$.

(C) Let $y = 3^{12} \times 2^8$.
Taking $\log_{10}$ on both sides:
$\log_{10} y = \log_{10} (3^{12} \times 2^8) = 12 \log_{10} 3 + 8 \log_{10} 2$.
Substituting the given values:
$\log_{10} y = 12(0.47712) + 8(0.30103)$.
$\log_{10} y = 5.72544 + 2.40824 = 8.13368$.
The number of digits in a number $y$ is given by $\lfloor \log_{10} y \rfloor + 1$.
Number of digits $= \lfloor 8.13368 \rfloor + 1 = 8 + 1 = 9$.

(A) We are given the sum $S = \sum\limits_{k = 1}^n \frac{1}{\log_{2^k}(a)}$.
Using the change of base property $\frac{1}{\log_b a} = \log_a b$,we have $\frac{1}{\log_{2^k}(a)} = \log_a(2^k)$.
Thus,$S = \sum\limits_{k = 1}^n \log_a(2^k)$.
Using the property $\log_a(x^y) = y \log_a x$,we get $S = \sum\limits_{k = 1}^n k \log_a 2$.
Since $\log_a 2$ is a constant,$S = (\log_a 2) \sum\limits_{k = 1}^n k$.
Using the sum formula $\sum\limits_{k = 1}^n k = \frac{n(n + 1)}{2}$,we obtain $S = \frac{n(n + 1)}{2} \log_a 2$.

(C) Given the equation $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$.
Since $\log_{7}(1) = 0$,we have $\log_{5}(\sqrt{x^2 + x + 5}) = 1$.
This implies $\sqrt{x^2 + x + 5} = 5^1 = 5$.
Squaring both sides,we get $x^2 + x + 5 = 25$.
Rearranging the terms,$x^2 + x - 20 = 0$.
Factoring the quadratic equation,$(x + 5)(x - 4) = 0$.
Thus,$x = 4$ or $x = -5$.
Checking the domain,for $\log_{5}(\sqrt{x^2 + x + 5})$ to be defined,$\sqrt{x^2 + x + 5} > 0$,which is satisfied for both values.
However,the original equation is $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$.
For $x = 4$,$\sqrt{16+4+5} = \sqrt{25} = 5$,and $\log_{5}(5) = 1$,$\log_{7}(1) = 0$.
For $x = -5$,$\sqrt{25-5+5} = \sqrt{25} = 5$,and $\log_{5}(5) = 1$,$\log_{7}(1) = 0$.
Given the options,$x = 4$ is the correct choice.

(B) We have ${\log _4}18 = {\log _{{2^2}}}({3^2} \times 2)$.
Using the property ${\log _{{a^n}}}{b^m} = {m \over n}{\log _a}b$,we get:
${\log _4}18 = {\log _{{2^2}}}({2 \times 3^2}) = {1 \over 2}{\log _2}(2 \times 3^2)$.
Applying the product rule ${\log _a}(xy) = {\log _a}x + {\log _a}y$:
${\log _4}18 = {1 \over 2}({\log _2}2 + {\log _2}{3^2}) = {1 \over 2}(1 + 2{\log _2}3) = {1 \over 2} + {\log _2}3$.
Since ${\log _2}3$ is an irrational number,${1 \over 2} + {\log _2}3$ is also an irrational number.
Therefore,the correct option is $B$.

(A) Let $x = \ln a, y = \ln b, z = \ln c$. Since $a, b, c \neq 1$,$x, y, z \neq 0$.
Given equation: $[(\frac{x}{y})(\frac{x}{z}) - 1] + [(\frac{y}{x})(\frac{y}{z}) - 1] + [(\frac{z}{x})(\frac{z}{y}) - 1] = 0$
$\Rightarrow \frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} - 3 = 0$
$\Rightarrow \frac{x^3 + y^3 + z^3}{xyz} = 3$
$\Rightarrow x^3 + y^3 + z^3 - 3xyz = 0$
Using the identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0$,we have either $x + y + z = 0$ or $x^2 + y^2 + z^2 - xy - yz - zx = 0$.
If $x^2 + y^2 + z^2 - xy - yz - zx = 0$,then $\frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$,which implies $x = y = z$,but $a, b, c$ are distinct.
Therefore,$x + y + z = 0$,which means $\ln a + \ln b + \ln c = 0$.
$\Rightarrow \ln(abc) = 0 = \ln 1$.
Thus,$abc = 1$.

(C) Given $a = \log_{12} 27 = \frac{\log 27}{\log 12} = \frac{3 \log 3}{\log 3 + 2 \log 2}$.
Rearranging for $\log 3$: $a(\log 3 + 2 \log 2) = 3 \log 3$ $\Rightarrow a \log 3 + 2a \log 2 = 3 \log 3$ $\Rightarrow 2a \log 2 = (3 - a) \log 3$ $\Rightarrow \log 3 = \frac{2a \log 2}{3 - a}$.
Now,evaluate $\log_6 16 = \frac{\log 16}{\log 6} = \frac{4 \log 2}{\log 2 + \log 3}$.
Substituting $\log 3$: $\frac{4 \log 2}{\log 2 + \frac{2a \log 2}{3 - a}} = \frac{4 \log 2}{\frac{(3 - a) \log 2 + 2a \log 2}{3 - a}} = \frac{4(3 - a) \log 2}{(3 - a + 2a) \log 2} = \frac{4(3 - a)}{3 + a}$.

(B) Let $\log_{16} x = y$. The equation becomes $y^2 - y + \log_{16} k = 0$.
This quadratic equation in $y$ will have exactly one solution for $x$ if the discriminant $D$ is equal to $0$.
$D = (-1)^2 - 4(1)(\log_{16} k) = 0$.
$1 - 4\log_{16} k = 0 \Rightarrow \log_{16} k = \frac{1}{4}$.
$k = 16^{1/4} = (2^4)^{1/4} = 2$.
Since $\log_{16} k$ must be defined,we require $k > 0$. Thus,$k = 2$ is the only valid real value.
Therefore,the number of real values of $k$ is $1$.

(A) Given the inequality: $\frac{1}{\log_3 \pi} + \frac{1}{\log_4 \pi} > x$
Using the property $\frac{1}{\log_a b} = \log_b a$,we get:
$\log_{\pi} 3 + \log_{\pi} 4 > x$
Using the property $\log_b m + \log_b n = \log_b (mn)$,we get:
$\log_{\pi} (3 \times 4) > x$
$\log_{\pi} 12 > x$
We know that $\pi \approx 3.14$,so $\pi^2 \approx 9.86$ and $\pi^3 \approx 31.00$.
Since $9.86 < 12 < 31.00$,it follows that $\pi^2 < 12 < \pi^3$.
Taking $\log_{\pi}$ on all sides:
$\log_{\pi} (\pi^2) < \log_{\pi} 12 < \log_{\pi} (\pi^3)$
$2 < \log_{\pi} 12 < 3$
Since $\log_{\pi} 12 > x$ and $\log_{\pi} 12 > 2$,the value of $x$ can be $2$.

(A) Given the inequality $\log _{1/\sqrt{2}} \sin x > 0$.
Since the base $1/\sqrt{2}$ is between $0$ and $1$,the inequality reverses when we remove the logarithm:
$0 < \sin x < (1/\sqrt{2})^0$
$0 < \sin x < 1$
We need to find the values of $x = k \cdot \frac{\pi}{4}$ for $k \in \mathbb{Z}$ in the interval $[0, 4\pi]$ such that $0 < \sin x < 1$.
The multiples of $\frac{\pi}{4}$ in $[0, 4\pi]$ are $0, \frac{\pi}{4}, \frac{2\pi}{4}, \frac{3\pi}{4}, \dots, \frac{16\pi}{4}$.
Values where $\sin x = 0$ or $\sin x = 1$ must be excluded.
For $x \in [0, 4\pi]$,$\sin x = 0$ at $x = 0, \pi, 2\pi, 3\pi, 4\pi$.
$\sin x = 1$ at $x = \frac{\pi}{2}, \frac{5\pi}{2}$.
Checking multiples of $\frac{\pi}{4}$:
$x = \frac{\pi}{4} (\sin > 0), \frac{2\pi}{4} (\sin = 1, \text{exclude}), \frac{3\pi}{4} (\sin > 0), \frac{4\pi}{4} (\sin = 0, \text{exclude}), \frac{5\pi}{4} (\sin < 0), \frac{6\pi}{4} (\sin < 0), \frac{7\pi}{4} (\sin < 0), \frac{8\pi}{4} (\sin = 0, \text{exclude}), \frac{9\pi}{4} (\sin > 0), \frac{10\pi}{4} (\sin = 1, \text{exclude}), \frac{11\pi}{4} (\sin > 0), \frac{12\pi}{4} (\sin = 0, \text{exclude}), \frac{13\pi}{4} (\sin < 0), \frac{14\pi}{4} (\sin < 0), \frac{15\pi}{4} (\sin < 0), \frac{16\pi}{4} (\sin = 0, \text{exclude})$.
The valid values are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}$.
There are $4$ such values.

(B) Given the inequality: $\log _{1/2}(x^2 - 6x + 12) \ge -2$ $(i)$
For the logarithm to be defined,the argument must be positive: $x^2 - 6x + 12 > 0$.
Since the discriminant $D = (-6)^2 - 4(1)(12) = 36 - 48 = -12 < 0$ and the leading coefficient is positive,$x^2 - 6x + 12 > 0$ for all $x \in \mathbb{R}$.
Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:
$x^2 - 6x + 12 \le (1/2)^{-2}$
$x^2 - 6x + 12 \le 4$
$x^2 - 6x + 8 \le 0$
$(x - 2)(x - 4) \le 0$
Solving the quadratic inequality,we get $2 \le x \le 4$.
Thus,the solution set is $x \in [2, 4]$.

(B) Given the inequality: $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$
First,note the domain condition for the logarithm: $x - 1 > 0 \Rightarrow x > 1$.
Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{\sqrt{2}}(x - 1) = \log_{2^{1/2}}(x - 1) = 2 \log_2(x - 1) = \log_2((x - 1)^2)$.
Substituting this back into the inequality: $2^{\log_2((x - 1)^2)} > x + 5$.
Using the identity $a^{\log_a(y)} = y$,we get: $(x - 1)^2 > x + 5$.
Expanding the expression: $x^2 - 2x + 1 > x + 5 \Rightarrow x^2 - 3x - 4 > 0$.
Factoring the quadratic: $(x - 4)(x + 1) > 0$.
The solution to this inequality is $x < -1$ or $x > 4$.
Considering the domain condition $x > 1$,the intersection of $x > 1$ and $(x < -1 \cup x > 4)$ is $x > 4$.
Thus,the set of real values is $(4, \infty )$.

(C) Given the inequality: $\log _{0.04}(x - 1) \ge \log _{0.2}(x - 1)$ $(i)$
For the logarithm to be defined,we must have $x - 1 > 0$,which implies $x > 1$.
Rewrite the base $0.04$ as $(0.2)^2$:
$\log _{(0.2)^2}(x - 1) \ge \log _{0.2}(x - 1)$
Using the property $\log _{a^n}(b) = \frac{1}{n} \log _a(b)$:
$\frac{1}{2} \log _{0.2}(x - 1) \ge \log _{0.2}(x - 1)$
Subtract $\frac{1}{2} \log _{0.2}(x - 1)$ from both sides:
$0 \ge \frac{1}{2} \log _{0.2}(x - 1)$
Since the base $0.2 < 1$,the inequality reverses when we remove the log:
$x - 1 \ge (0.2)^0$
$x - 1 \ge 1$
$x \ge 2$
Combining with the domain $x > 1$,we get $x \in [2, \infty)$.

(A) Given the inequality: $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ ... $(i)$
For the logarithm to be defined,the argument must be positive:
$\frac{x + 2}{x} > 0 \Rightarrow x \in (-\infty, -2) \cup (0, \infty)$
Since the base $0.2 < 1$,the inequality sign reverses when removing the logarithm:
$\frac{x + 2}{x} \ge (0.2)^1$
$\frac{x + 2}{x} \ge 0.2$
$\frac{x + 2}{x} - 0.2 \ge 0$
$\frac{x + 2 - 0.2x}{x} \ge 0$
$\frac{0.8x + 2}{x} \ge 0$
$\frac{4x + 10}{5x} \ge 0$
$\frac{2(2x + 5)}{5x} \ge 0$
Using the wavy curve method,the solution to $\frac{2x + 5}{x} \ge 0$ is $x \in (-\infty, -2.5] \cup (0, \infty)$.
Combining this with the domain condition $x \in (-\infty, -2) \cup (0, \infty)$,the intersection is $x \in (-\infty, -2.5] \cup (0, \infty)$.

(B) Given that $x = \log_{b}a$,$y = \log_{c}b$,and $z = \log_{a}c$.
We need to find the product $xyz$.
Using the change of base formula,$\log_{n}m = \frac{\log_{k}m}{\log_{k}n}$,we can write:
$x = \frac{\log a}{\log b}$,$y = \frac{\log b}{\log c}$,and $z = \frac{\log c}{\log a}$.
Now,multiply these values:
$xyz = \left(\frac{\log a}{\log b}\right) \times \left(\frac{\log b}{\log c}\right) \times \left(\frac{\log c}{\log a}\right)$.
Canceling the common terms in the numerator and denominator,we get:
$xyz = 1$.

(B) Let the given expression be $E = \log_2(\log_3(\dots(\log_{100}(100^{99^{98^{\dots^{2^1}}})))\dots))}$.
Using the property $\log_a(a^x) = x$,we simplify from the innermost logarithm:
$\log_{100}(100^{99^{98^{\dots^{2^1}}}}) = 99^{98^{\dots^{2^1}}}$.
Substituting this back,we get:
$E = \log_2(\log_3(\dots(\log_{99}(99^{98^{\dots^{2^1}}})))\dots))$.
Continuing this process,the expression simplifies step by step:
$\log_{99}(99^{98^{\dots^{2^1}}}) = 98^{\dots^{2^1}}$.
This pattern continues until we reach:
$\log_3(3^{2^1}) = 2^1 = 2$.
Finally,we have $E = \log_2(2) = 1$.

(C) Given equations are:
$a^x = bc$,$b^y = ca$,$c^z = ab$
Taking logarithm on both sides for each equation (assuming base $a, b, c$ are consistent):
$x = \log_a(bc) = \log_a b + \log_a c$
$y = \log_b(ca) = \log_b c + \log_b a$
$z = \log_c(ab) = \log_c a + \log_c b$
Adding $1$ to each side:
$x+1 = \log_a b + \log_a c + 1 = \log_a(abc)$
$y+1 = \log_b c + \log_b a + 1 = \log_b(abc)$
$z+1 = \log_c a + \log_c b + 1 = \log_c(abc)$
Taking reciprocals:
$\frac{1}{x+1} = \log_{abc} a$,$\frac{1}{y+1} = \log_{abc} b$,$\frac{1}{z+1} = \log_{abc} c$
Summing these:
$\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc}(abc) = 1$
Alternatively,if $a=b=c=2$,then $2^x = 4 \Rightarrow x=2$,$2^y = 4 \Rightarrow y=2$,$2^z = 4 \Rightarrow z=2$.
Then $xyz = 2 \times 2 \times 2 = 8$.
Checking options: $x+y+z+2 = 2+2+2+2 = 8$.
Thus,$xyz = x+y+z+2$.

(C) Given $x = \log_{3} 5$ and $y = \log_{17} 25 = 2 \log_{17} 5$.
Taking the reciprocal of $y$,we get $\frac{1}{y} = \frac{1}{2 \log_{17} 5} = \frac{1}{2} \log_{5} 17 = \log_{5} (17^{1/2}) = \log_{5} \sqrt{17}$.
Taking the reciprocal of $x$,we get $\frac{1}{x} = \log_{5} 3 = \log_{5} \sqrt{9}$.
Since $\sqrt{17} > \sqrt{9}$,it follows that $\log_{5} \sqrt{17} > \log_{5} \sqrt{9}$.
Therefore,$\frac{1}{y} > \frac{1}{x}$,which implies $x > y$.

(D) Given the equation $\log_{2}(x + 5) = 6 - x$.
Let $f(x) = \log_{2}(x + 5)$ and $g(x) = 6 - x$.
$f(x)$ is a strictly increasing function for $x > -5$.
$g(x)$ is a strictly decreasing function.
At $x = 3$,$f(3) = \log_{2}(3 + 5) = \log_{2}(8) = 3$ and $g(3) = 6 - 3 = 3$.
Since $f(x)$ is strictly increasing and $g(x)$ is strictly decreasing,they intersect at exactly one point.
Therefore,the number of solutions is $1$.

(B) We need to compare $\log_{20} 3$ with given intervals.
Since $20^{1/3} = \sqrt[3]{20}$ and $2^3 = 8$,$3^3 = 27$,we know $2 < \sqrt[3]{20} < 3$,so $20^{1/3} < 3$.
Also,$20^{1/2} = \sqrt{20} \approx 4.47$,so $3 < 20^{1/2}$.
Thus,$20^{1/3} < 3 < 20^{1/2}$.
Taking $\log_{20}$ on all sides,we get $\frac{1}{3} < \log_{20} 3 < \frac{1}{2}$.
Therefore,$\log_{20} 3 \in (1/3, 1/2)$.

(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we have $\log_{3^{1/k}} x = k \log_3 x$.
Substituting this into the equation:
$2 \log_3 x + 4 \log_3 x + 6 \log_3 x + \dots + 16 \log_3 x = 36$
Factor out $\log_3 x$:
$(\log_3 x) (2 + 4 + 6 + \dots + 16) = 36$
The sum of the arithmetic progression $2 + 4 + \dots + 16$ with $n = 8$ terms is $\frac{8}{2}(2 + 16) = 4 \times 18 = 72$.
So,$(\log_3 x) \times 72 = 36$
$\log_3 x = \frac{36}{72} = \frac{1}{2}$
$x = 3^{1/2} = \sqrt{3}$.

(A) The given equation is $\log _a x + \log _{a^{1/2}} x + \log _{a^{1/3}} x + \dots + \log _{a^{1/n}} x = \frac{n(n+1)}{2}$.
Using the property $\log _{a^k} x = \frac{1}{k} \log_a x$,we get:
$\log_a x + 2 \log_a x + 3 \log_a x + \dots + n \log_a x = \frac{n(n+1)}{2}$.
$\log_a x (1 + 2 + 3 + \dots + n) = \frac{n(n+1)}{2}$.
Since the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$,we have:
$\log_a x \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2}$.
$\log_a x = 1$.
Therefore,$x = a^1 = a$.

(D) Given equation: $2^{x + 2} \cdot (3^3)^{x/(x - 1)} = 3^2$
Taking $\log$ on both sides:
$(x + 2)\log 2 + \frac{3x}{x - 1}\log 3 = 2\log 3$
$(x + 2)\log 2 = 2\log 3 - \frac{3x}{x - 1}\log 3$
$(x + 2)\log 2 = \log 3 \left( 2 - \frac{3x}{x - 1} \right)$
$(x + 2)\log 2 = \log 3 \left( \frac{2x - 2 - 3x}{x - 1} \right)$
$(x + 2)\log 2 = \log 3 \left( \frac{-x - 2}{x - 1} \right)$
$(x + 2)\log 2 = -\log 3 \left( \frac{x + 2}{x - 1} \right)$
$(x + 2) \left( \log 2 + \frac{\log 3}{x - 1} \right) = 0$
Case $1$: $x + 2 = 0 \implies x = -2$
Case $2$: $\log 2 + \frac{\log 3}{x - 1} = 0 \implies \frac{\log 3}{x - 1} = -\log 2$
$x - 1 = -\frac{\log 3}{\log 2} \implies x = 1 - \frac{\log 3}{\log 2}$

(D) Let $t = \log_2 x$. The given equation is $t + \frac{1}{t} = \frac{10}{3}$.
Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.
Factoring the quadratic,we have $(3t - 1)(t - 3) = 0$.
Thus,$t = 3$ or $t = \frac{1}{3}$.
Since $x \neq y$,we assign $\log_2 x = 3$ and $\log_2 y = \frac{1}{3}$ (or vice versa).
This gives $x = 2^3 = 8$ and $y = 2^{1/3} = \sqrt[3]{2}$.
Therefore,$x + y = 8 + \sqrt[3]{2}$.

(B) Given the equation: $\log_e x + \log_e(1 + x) = 0$
Using the logarithmic property $\log_e a + \log_e b = \log_e(ab)$,we get:
$\log_e(x(1 + x)) = 0$
Applying the definition of logarithm,where $\log_e y = 0$ implies $y = e^0 = 1$:
$x(1 + x) = 1$
$x^2 + x = 1$
$x^2 + x - 1 = 0$
Thus,the correct option is $B$.

(A) Given the equation: $\log_{4}\{\log_{2}(\sqrt{x + 8} - \sqrt{x})\} = 0$
By the definition of logarithm,$\log_{b}(a) = c \implies a = b^c$. Thus:
$\log_{2}(\sqrt{x + 8} - \sqrt{x}) = 4^0 = 1$
Again,applying the definition of logarithm:
$\sqrt{x + 8} - \sqrt{x} = 2^1 = 2$
Rearranging the terms:
$\sqrt{x + 8} = 2 + \sqrt{x}$
Squaring both sides:
$x + 8 = (2 + \sqrt{x})^2$
$x + 8 = 4 + x + 4\sqrt{x}$
Subtracting $x$ from both sides:
$8 = 4 + 4\sqrt{x}$
$4 = 4\sqrt{x}$
$1 = \sqrt{x}$
Squaring both sides again:
$x = 1$
Checking the domain: For $\sqrt{x+8} - \sqrt{x}$ to be defined,$x \ge 0$. For $\log_{2}(\sqrt{x+8} - \sqrt{x})$ to be defined,$\sqrt{x+8} - \sqrt{x} > 0$,which is true for $x=1$ as $\sqrt{9} - 1 = 3 - 1 = 2 > 0$.

(B) Given equation: $\log_{4}(x - 1) = \log_{2}(x - 3)$
Using the base change formula $\log_{a^n}(b) = \frac{1}{n} \log_{a}(b)$,we have $\log_{4}(x - 1) = \frac{1}{2} \log_{2}(x - 1)$.
So,$\frac{1}{2} \log_{2}(x - 1) = \log_{2}(x - 3)$
$\log_{2}(x - 1) = 2 \log_{2}(x - 3)$
$\log_{2}(x - 1) = \log_{2}((x - 3)^2)$
$x - 1 = (x - 3)^2$
$x - 1 = x^2 - 6x + 9$
$x^2 - 7x + 10 = 0$
$(x - 5)(x - 2) = 0$
Thus,$x = 5$ or $x = 2$.
Check the domain: For $\log_{2}(x - 3)$ to be defined,$x - 3 > 0$,so $x > 3$.
For $x = 2$,$x - 3 = -1$,which is not allowed.
For $x = 5$,$x - 3 = 2 > 0$,which is valid.
Therefore,there is only $1$ solution.

(D) For the given equation to be meaningful,we must have $x > 0$. Taking the logarithm base $2$ on both sides:
$(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \log_2 x = \log_2 \sqrt{2} = \frac{1}{2}$.
Let $t = \log_2 x$. Then the equation becomes:
$(\frac{3}{4}t^2 + t - \frac{5}{4}) t = \frac{1}{2}$.
Multiplying by $4$:
$(3t^2 + 4t - 5) t = 2 \Rightarrow 3t^3 + 4t^2 - 5t - 2 = 0$.
By testing roots,we find $(t - 1)$ is a factor:
$(t - 1)(3t^2 + 7t + 2) = 0 \Rightarrow (t - 1)(3t + 1)(t + 2) = 0$.
Thus,$t = 1, -2, -1/3$.
Since $t = \log_2 x$,we have $x = 2^1 = 2$,$x = 2^{-2} = 1/4$,and $x = 2^{-1/3} = 1/\sqrt[3]{2}$.
All three solutions are real,and $1/\sqrt[3]{2}$ is irrational. Therefore,all statements $(A)$,$(B)$,and $(C)$ are correct.

(C) Given equation: $\log(-2x) = 2\log(x+1)$.
For the equation to be defined,the arguments of the logarithm must be positive:
$-2x > 0 \Rightarrow x < 0$ and $x+1 > 0 \Rightarrow x > -1$.
Thus,the domain is $x \in (-1, 0)$.
Applying logarithm properties: $\log(-2x) = \log((x+1)^2)$.
Equating the arguments: $-2x = (x+1)^2$.
$-2x = x^2 + 2x + 1 \Rightarrow x^2 + 4x + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$.
Check the values against the domain $x \in (-1, 0)$:
$x_1 = -2 + \sqrt{3} \approx -2 + 1.732 = -0.268$ (This is in the domain).
$x_2 = -2 - \sqrt{3} \approx -2 - 1.732 = -3.732$ (This is not in the domain).
Therefore,there is only $1$ valid root.

(C) The logarithmic function $\log_{a} x$ is defined if and only if the base $a > 0, a \neq 1$ and the argument $x > 0$.
Therefore,$\log_{a} x$ is defined for all positive real values of $x$.

(A) Given the expression $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ where $n = (1999)!$.
Using the property of logarithms $\log_b a + \log_b c = \log_b (ac)$,we can write:
$\sum\limits_{x = 1}^{1999} {{\log }_n x} = \log_n 1 + \log_n 2 + \dots + \log_n 1999$
$= \log_n (1 \times 2 \times 3 \times \dots \times 1999)$
$= \log_n (1999)!$
Since $n = (1999)!$,the expression becomes $\log_{(1999)!} (1999)!$.
Using the property $\log_a a = 1$,we get $1$.

(B) The function $f(x) = \log(x^2)$ is defined for all $x \in \mathbb{R} \setminus \{0\}$.
Using the logarithmic property $\log(a^n) = n \log a$,we have $\log(x^2) = 2 \log |x|$.
Note that $\log x$ is only defined for $x > 0$,whereas $\log(x^2)$ is defined for both positive and negative values of $x$ (excluding $0$).
Therefore,the equivalent function is $2 \log |x|$.

(C) Given $y = 2^{1/\log_x 4}$.
Taking logarithm on both sides with base $e$,we get $\ln y = \frac{1}{\log_x 4} \ln 2$.
Using the change of base formula $\log_x 4 = \frac{\ln 4}{\ln x}$,we have $\ln y = \frac{\ln 2}{\frac{\ln 4}{\ln x}} = \frac{\ln 2 \cdot \ln x}{\ln 4}$.
Since $\ln 4 = \ln(2^2) = 2 \ln 2$,the equation becomes $\ln y = \frac{\ln 2 \cdot \ln x}{2 \ln 2}$.
Simplifying,we get $\ln y = \frac{\ln x}{2}$.
Multiplying by $2$,we have $2 \ln y = \ln x$.
Using the property $n \ln a = \ln(a^n)$,we get $\ln(y^2) = \ln x$.
Therefore,$x = y^2$.

(B) Using the change of base formula,$\frac{1}{\log_a n} = \log_n a$.
Given expression: $\log_n 2 + \log_n 3 + ... + \log_n 1000$.
This simplifies to $\log_n (2 \times 3 \times ... \times 1000)$.
Since $n = 1000!$,the expression becomes $\log_{1000!} (1000!)$.
Therefore,the result is $1$.

(B) Given the equation $x^{\log_x(1-x)^2} = 9$.
Using the property $x^{\log_x(A)} = A$,we have $(1-x)^2 = 9$.
Expanding the equation: $1 - 2x + x^2 = 9$.
Rearranging terms: $x^2 - 2x - 8 = 0$.
Factoring the quadratic: $(x - 4)(x + 2) = 0$.
This gives $x = 4$ or $x = -2$.
However,for the term $\log_x(1-x)^2$ to be defined,the base $x$ must satisfy $x > 0$ and $x \neq 1$.
Checking $x = 4$: Base is $4 > 0$ and $4 \neq 1$. Also,$(1-4)^2 = 9 > 0$. This is a valid solution.
Checking $x = -2$: Base is $-2$,which is not allowed $(x > 0)$.
Thus,the only valid solution is $x = 4$.

(A) Given the equation: ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$
Applying the definition of logarithm,we get: ${\log _2}(\sqrt {x + 8} - \sqrt x ) = {4^0} = 1$
Again,applying the definition of logarithm: $\sqrt {x + 8} - \sqrt x = {2^1} = 2$
Squaring both sides: $(\sqrt {x + 8} - \sqrt x)^2 = {2^2}$
$x + 8 + x - 2\sqrt {x(x + 8)} = 4$
$2x + 8 - 2\sqrt {x^2 + 8x} = 4$
$2x + 4 = 2\sqrt {x^2 + 8x}$
Dividing by $2$: $x + 2 = \sqrt {x^2 + 8x}$
Squaring both sides again: $(x + 2)^2 = x^2 + 8x$
$x^2 + 4x + 4 = x^2 + 8x$
$4 = 4x$
$x = 1$

(B) Given the equation: $\log_4(x - 1) = \log_2(x - 3)$.
First,note the domain constraints: $x - 1 > 0 \implies x > 1$ and $x - 3 > 0 \implies x > 3$. Thus,$x > 3$.
Using the change of base formula,$\log_4(x - 1) = \frac{\log_2(x - 1)}{\log_2(4)} = \frac{\log_2(x - 1)}{2}$.
Substituting this into the equation: $\frac{\log_2(x - 1)}{2} = \log_2(x - 3)$.
$\log_2(x - 1) = 2 \log_2(x - 3) = \log_2((x - 3)^2)$.
Equating the arguments: $x - 1 = (x - 3)^2$.
$x - 1 = x^2 - 6x + 9$.
$x^2 - 7x + 10 = 0$.
$(x - 2)(x - 5) = 0$.
So,$x = 2$ or $x = 5$.
Checking the domain $x > 3$: $x = 2$ is rejected,and $x = 5$ is accepted.
Therefore,there is only $1$ solution.