Logarithms Questions in English

(B) Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we can rewrite the expression as:
$\log _{n} 2 + \log _{n} 3 + \log _{n} 4 + \ldots + \log _{n} 2020$
Using the property $\log _{b} x + \log _{b} y = \log _{b} (xy)$,we get:
$\log _{n} (2 \times 3 \times 4 \times \ldots \times 2020)$
Since $n = (2020)!$,the expression becomes:
$\log _{n} (n) = 1$

(D) The given expression is $P = \log(\sin 1^{\circ}) \cdot \log(\sin 2^{\circ}) \cdot \dots \cdot \log(\sin 90^{\circ}) \cdot \dots \cdot \log(\sin 179^{\circ})$.
We know that $\sin 90^{\circ} = 1$.
Therefore,the term $\log(\sin 90^{\circ}) = \log(1) = 0$.
Since the expression is a product of terms including $\log(\sin 90^{\circ})$,the entire product becomes $0$ because any number multiplied by $0$ is $0$.
Thus,the value is $0$.

(D) Taking $\log_2$ on both sides,we get:
$(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \cdot \log_2 x = \log_2(2^{1/2})$
Let $y = \log_2 x$. Then the equation becomes:
$(\frac{3}{4}y^2 + y - \frac{5}{4})y = \frac{1}{2}$
Multiplying by $4$:
$(3y^2 + 4y - 5)y = 2$
$3y^3 + 4y^2 - 5y - 2 = 0$
By testing values,$y = 1$ is a root since $3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0$.
Dividing by $(y - 1)$,we get $(y - 1)(3y^2 + 7y + 2) = 0$.
Further factoring: $(y - 1)(3y + 1)(y + 2) = 0$.
The roots are $y = 1, y = -1/3, y = -2$.
Since $y = \log_2 x$,we have $x = 2^1, x = 2^{-1/3}, x = 2^{-2}$.
All three values are real and positive,so there are exactly three real solutions.

(D) We use the property $\log a + \log b = \log (ab)$.
$\log (\cosh 3 + \sinh 3) + \log (\cosh 3 - \sinh 3) = \log ((\cosh 3 + \sinh 3)(\cosh 3 - \sinh 3))$
$= \log (\cosh^2 3 - \sinh^2 3)$
Since the identity for hyperbolic functions is $\cosh^2 x - \sinh^2 x = 1$,we have:
$= \log (1)$
$= 0$

(A) Given,$a^x = b^y = c^z = d^w = k$ (let).
Then,$x = \log_a k$,$y = \log_b k$,$z = \log_c k$,$w = \log_d k$.
Taking reciprocals,$\frac{1}{x} = \log_k a$,$\frac{1}{y} = \log_k b$,$\frac{1}{z} = \log_k c$,$\frac{1}{w} = \log_k d$.
We need to evaluate $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$.
Substituting the values,we get $x\left(\log_k b + \log_k c + \log_k d\right) = x \log_k(bcd)$.
Since $x = \log_a k$,the expression becomes $\log_a k \cdot \log_k(bcd)$.
Using the change of base formula $\log_a k \cdot \log_k(bcd) = \log_a(bcd)$.

(D) Given expression: $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$
Simplify the denominator:
$10^{-4} a+10^{-3} b+10^{-2} c = \frac{a}{10^4} + \frac{b}{10^3} + \frac{c}{10^2} = \frac{a+10b+10^2c}{10^4}$
Substitute back into the logarithm:
$\log _{10}\left(\frac{a+10 b+10^2 c}{\frac{a+10 b+10^2 c}{10^4}}\right) = \log _{10}(10^4)$
Using the property $\log_b(b^x) = x$:
$\log _{10}(10^4) = 4$

(D) Given that,$\log _{27}(\log _3 x) = \frac{1}{3}$.
Using the definition of logarithm,$\log _a b = c \Rightarrow b = a^c$,we get:
$\log _3 x = (27)^{1/3}$.
Since $27 = 3^3$,we have $(27)^{1/3} = (3^3)^{1/3} = 3^1 = 3$.
So,$\log _3 x = 3$.
Again,using the definition of logarithm,$x = 3^3$.
Therefore,$x = 27$.

(A) Given the equation: $3 \sinh (2 \theta) = 13 - 3 e^{2 \theta}$
Substitute $\sinh (2 \theta) = \frac{e^{2 \theta} - e^{-2 \theta}}{2}$:
$3 \left( \frac{e^{2 \theta} - e^{-2 \theta}}{2} \right) = 13 - 3 e^{2 \theta}$
Let $x = e^{2 \theta}$. Since $e^{2 \theta} > 0$,we have $x > 0$.
$\frac{3}{2} (x - \frac{1}{x}) = 13 - 3x$
$3(x^2 - 1) = 2x(13 - 3x)$
$3x^2 - 3 = 26x - 6x^2$
$9x^2 - 26x - 3 = 0$
Factoring the quadratic: $9x^2 - 27x + x - 3 = 0$
$9x(x - 3) + 1(x - 3) = 0$
$(9x + 1)(x - 3) = 0$
Since $x > 0$,we must have $x = 3$.
$e^{2 \theta} = 3$
Taking the natural logarithm on both sides:
$2 \theta = \log 3$
$\theta = \frac{1}{2} \log 3$

(D) Let $I = \frac{\cosh \alpha}{1-\tanh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha}$.
Substitute $\tanh \alpha = \frac{\sinh \alpha}{\cosh \alpha}$ and $\coth \alpha = \frac{\cosh \alpha}{\sinh \alpha}$:
$I = \frac{\cosh \alpha}{1 - \frac{\sinh \alpha}{\cosh \alpha}} + \frac{\sinh \alpha}{1 - \frac{\cosh \alpha}{\sinh \alpha}}$
$I = \frac{\cosh^2 \alpha}{\cosh \alpha - \sinh \alpha} + \frac{\sinh^2 \alpha}{\sinh \alpha - \cosh \alpha}$
$I = \frac{\cosh^2 \alpha - \sinh^2 \alpha}{\cosh \alpha - \sinh \alpha}$
Since $\cosh^2 \alpha - \sinh^2 \alpha = 1$,we have $I = \frac{1}{\cosh \alpha - \sinh \alpha}$.
Using the identity $\cosh \alpha - \sinh \alpha = e^{-\alpha}$,we get $I = \frac{1}{e^{-\alpha}} = e^{\alpha}$.
Given $\alpha = \log_e(2+\sqrt{3})$,then $e^{\alpha} = 2+\sqrt{3}$.

(B) We know the logarithmic forms of inverse hyperbolic functions:
$\sec h^{-1} x = \log \left(\frac{1+\sqrt{1-x^2}}{x}\right)$
$\tan h^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$
$\sin h^{-1} x = \log \left(x+\sqrt{x^2+1}\right)$
For $x = \frac{1}{2}$:
$\sec h^{-1} \frac{1}{2} = \log \left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log (2+\sqrt{3})$
$\tan h^{-1} \frac{1}{2} = \frac{1}{2} \log \left(\frac{3/2}{1/2}\right) = \frac{1}{2} \log 3 = \log \sqrt{3}$
$\sin h^{-1} \frac{1}{2} = \log \left(\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \log \left(\frac{1+\sqrt{5}}{2}\right)$
Now,the expression becomes:
$e^{\log (2+\sqrt{3}) + \log \sqrt{3} + \log \left(\frac{1+\sqrt{5}}{2}\right)}$
$= e^{\log \left((2+\sqrt{3}) \cdot \sqrt{3} \cdot \frac{1+\sqrt{5}}{2}\right)}$
$= (2\sqrt{3}+3) \cdot \left(\frac{1+\sqrt{5}}{2}\right)$
$= \frac{2\sqrt{3} + 2\sqrt{15} + 3 + 3\sqrt{5}}{2}$
$= \frac{3 + 2\sqrt{3} + 3\sqrt{5} + 2\sqrt{15}}{2}$

(A) We use the logarithmic definitions of inverse hyperbolic functions: $\tanh^{-1}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$ and $\operatorname{coth}^{-1}(x) = \frac{1}{2} \ln(\frac{x+1}{x-1})$.
For $\tanh^{-1}(\frac{1}{2})$:
$\tanh^{-1}(\frac{1}{2}) = \frac{1}{2} \ln(\frac{1+1/2}{1-1/2}) = \frac{1}{2} \ln(\frac{3/2}{1/2}) = \frac{1}{2} \ln(3) = \log \sqrt{3}$.
For $\operatorname{coth}^{-1}(3)$:
$\operatorname{coth}^{-1}(3) = \frac{1}{2} \ln(\frac{3+1}{3-1}) = \frac{1}{2} \ln(\frac{4}{2}) = \frac{1}{2} \ln(2) = \log \sqrt{2}$.
Adding these results:
$\log \sqrt{3} + \log \sqrt{2} = \log(\sqrt{3} \cdot \sqrt{2}) = \log \sqrt{6}$.

(D) We know that $\operatorname{Sinh}^{-1}(y) = \log(y + \sqrt{y^2+1})$.
Let $y = \frac{a}{\sqrt{1-a^2}}$.
Then $y^2+1 = \frac{a^2}{1-a^2} + 1 = \frac{a^2+1-a^2}{1-a^2} = \frac{1}{1-a^2}$.
So,$\sqrt{y^2+1} = \frac{1}{\sqrt{1-a^2}}$ (since $a < 1$).
Thus,$\operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log\left(\frac{a+1}{\sqrt{1-a^2}}\right) = \log\left(\frac{a+1}{\sqrt{(1-a)(1+a)}}\right) = \log\left(\sqrt{\frac{1+a}{1-a}}\right) = \frac{1}{2} \log\left(\frac{1+a}{1-a}\right)$.
Given equation: $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Substituting the derived value: $2 \times \frac{1}{2} \log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Therefore,$\log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Comparing both sides,we get $x = a$.

(D) The identity is given by $\tanh^{-1} x = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} \right)$.
For the logarithmic function $\log_e(u)$ to be defined,we must have $u > 0$.
Therefore,we require $\frac{1+x}{1-x} > 0$.
Let $f(x) = \frac{1+x}{1-x}$. The critical points are $x = -1$ and $x = 1$.
Testing the intervals:
For $x < -1$,$f(x) < 0$.
For $-1 < x < 1$,$f(x) > 0$.
For $x > 1$,$f(x) < 0$.
At $x = 1$ and $x = -1$,the expression is undefined.
Thus,the identity is valid only for $x \in (-1, 1)$.

(B) Given $t=\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)$.
Using the logarithmic forms $\operatorname{sech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1+x^2}}{x}\right)$:
$t=\ln\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right)-\ln\left(\frac{1+\sqrt{1+(3/k)^2}}{3/k}\right)$
$t=\ln(2+\sqrt{3})-\ln\left(\frac{k+\sqrt{k^2+9}}{3}\right)$.
Given $3e^t=2+\sqrt{3}$,we have $e^t=\frac{2+\sqrt{3}}{3}$.
Taking the natural logarithm on both sides: $t=\ln\left(\frac{2+\sqrt{3}}{3}\right)$.
Equating the two expressions for $t$:
$\ln\left(\frac{2+\sqrt{3}}{3}\right)=\ln\left(\frac{2+\sqrt{3}}{\frac{k+\sqrt{k^2+9}}{3}}\right)$.
This implies $\frac{k+\sqrt{k^2+9}}{3}=3$,so $k+\sqrt{k^2+9}=9$.
$\sqrt{k^2+9}=9-k$.
Squaring both sides: $k^2+9=81+k^2-18k$.
$18k=72$,which gives $k=4$.

(B) We have,$\log (9+3 \sqrt{2}(2+\sqrt{5})+4 \sqrt{5})$
$= \log (9+6 \sqrt{2}+3 \sqrt{10}+4 \sqrt{5})$
$= \log ((3+\sqrt{10})(3+2 \sqrt{2}))$
$= \log (3+\sqrt{10}) + \log (3+\sqrt{8})$
$= \log (3+\sqrt{3^2+1}) + \log (3+\sqrt{3^2-1})$
Using the identities $\sinh^{-1}(x) = \log(x + \sqrt{x^2+1})$ and $\cosh^{-1}(x) = \log(x + \sqrt{x^2-1})$ for $x \ge 1$,we get:
$= \sinh^{-1} 3 + \cosh^{-1} 3$

(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.
For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\log(2+\sqrt{5})} - e^{-\log(2+\sqrt{5})}}{2} = \frac{(2+\sqrt{5}) - \frac{1}{2+\sqrt{5}}}{2}$.
Since $\frac{1}{2+\sqrt{5}} = \sqrt{5}-2$,we have $\frac{(2+\sqrt{5}) - (\sqrt{5}-2)}{2} = \frac{4}{2} = 2$.
For the second term: $\cosh[\log(2+\sqrt{3})] = \frac{e^{\log(2+\sqrt{3})} + e^{-\log(2+\sqrt{3})}}{2} = \frac{(2+\sqrt{3}) + \frac{1}{2+\sqrt{3}}}{2}$.
Since $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$,we have $\frac{(2+\sqrt{3}) + (2-\sqrt{3})}{2} = \frac{4}{2} = 2$.
Adding these results: $2 + 2 = 4$.

(A) We use the logarithmic definitions of inverse hyperbolic functions: $\tanh^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ for $|x| < 1$ and $\coth^{-1}(x) = \frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$ for $|x| > 1$.
Substituting $x = \frac{1}{3}$ in $\tanh^{-1}(x)$: $\tanh^{-1}\left(\frac{1}{3}\right) = \frac{1}{2} \log \left(\frac{1 + 1/3}{1 - 1/3}\right) = \frac{1}{2} \log \left(\frac{4/3}{2/3}\right) = \frac{1}{2} \log(2)$.
Substituting $x = 2$ in $\coth^{-1}(x)$: $\coth^{-1}(2) = \frac{1}{2} \log \left(\frac{2+1}{2-1}\right) = \frac{1}{2} \log(3)$.
Adding these results: $\frac{1}{2} \log(2) + \frac{1}{2} \log(3) = \frac{1}{2} \log(2 \times 3) = \frac{1}{2} \log(6) = \log(6^{1/2}) = \log \sqrt{6}$.

(D) Given,$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right)$.
We know that the inverse hyperbolic cosecant function is defined as $\operatorname{cosech}^{-1} y = \log \left( \frac{1}{y} + \sqrt{\frac{1}{y^2} + 1} \right)$.
Comparing this with the given equation,we get $x = \operatorname{cosech}^{-1} y$.
Taking the hyperbolic cosecant on both sides,we get $y = \operatorname{cosech} x$.

(C) We know that $\operatorname{sech}^{-1}(x) = \log_e \left( \frac{1+\sqrt{1-x^2}}{x} \right)$ and $\operatorname{cosech}^{-1}(x) = \log_e \left( \frac{1}{x} + \sqrt{\frac{1}{x^2}+1} \right)$.
Given equation: $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$.
First,calculate $\operatorname{sech}^{-1}\left(\frac{1}{2}\right) = \log_e \left( \frac{1+\sqrt{1-(1/2)^2}}{1/2} \right) = \log_e \left( 2(1+\sqrt{3/4}) \right) = \log_e (2+\sqrt{3})$.
Next,calculate $\operatorname{cosech}^{-1}\left(\frac{3}{4}\right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{16}{9}+1} \right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{25}{9}} \right) = \log_e \left( \frac{4}{3} + \frac{5}{3} \right) = \log_e(3)$.
Substituting these into the equation: $\log_e(2+\sqrt{3}) - \log_e(3) = \log_e k$.
$\log_e \left( \frac{2+\sqrt{3}}{3} \right) = \log_e k$.
So,$k = \frac{2+\sqrt{3}}{3}$,which implies $3k - 2 = \sqrt{3}$.
Squaring both sides: $(3k-2)^2 = 3$.
$9k^2 - 12k + 4 = 3$.
$9k^2 - 12k + 1 = 0$.

(B) We use the logarithmic definitions of inverse hyperbolic functions:
$\operatorname{coth}^{-1} x = \frac{1}{2} \log_e \left(\frac{x+1}{x-1}\right)$ for $|x| > 1$.
$\tanh^{-1} x = \frac{1}{2} \log_e \left(\frac{1+x}{1-x}\right)$ for $|x| < 1$.
$\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 + \sqrt{1+x^2}}{x}\right)$ for $x > 0$ and $\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 - \sqrt{1+x^2}}{x}\right)$ for $x < 0$.
Step $1$: Calculate each term.
$\operatorname{coth}^{-1}(3) = \frac{1}{2} \log_e \left(\frac{3+1}{3-1}\right) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$.
$\tanh^{-1}\left(\frac{1}{3}\right) = \frac{1}{2} \log_e \left(\frac{1+1/3}{1-1/3}\right) = \frac{1}{2} \log_e \left(\frac{4/3}{2/3}\right) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$.
$\operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \left(\frac{1 - \sqrt{1+(-\sqrt{3})^2}}{-\sqrt{3}}\right) = \log_e \left(\frac{1 - \sqrt{4}}{-\sqrt{3}}\right) = \log_e \left(\frac{1-2}{-\sqrt{3}}\right) = \log_e \left(\frac{-1}{-\sqrt{3}}\right) = \log_e \left(\frac{1}{\sqrt{3}}\right) = -\log_e \sqrt{3}$.
Step $2$: Combine the terms.
$\operatorname{coth}^{-1}(3) + \tanh^{-1}\left(\frac{1}{3}\right) - \operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} - (-\log_e \sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} + \log_e \sqrt{3} = \log_e (\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}) = \log_e (2\sqrt{3})$.

(C) We use the logarithmic forms of inverse hyperbolic functions: $\cosh ^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$ and $\sinh ^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$.
Given $\cosh ^{-1}\left(\frac{5}{3}\right) + \sinh ^{-1}\left(\frac{3}{4}\right) = k$.
Substituting the logarithmic forms:
$\ln\left(\frac{5}{3} + \sqrt{\left(\frac{5}{3}\right)^2 - 1}\right) + \ln\left(\frac{3}{4} + \sqrt{\left(\frac{3}{4}\right)^2 + 1}\right) = k$
$\ln\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}\right) + \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}\right) = k$
$\ln\left(\frac{5}{3} + \sqrt{\frac{16}{9}}\right) + \ln\left(\frac{3}{4} + \sqrt{\frac{25}{16}}\right) = k$
$\ln\left(\frac{5}{3} + \frac{4}{3}\right) + \ln\left(\frac{3}{4} + \frac{5}{4}\right) = k$
$\ln\left(\frac{9}{3}\right) + \ln\left(\frac{8}{4}\right) = k$
$\ln(3) + \ln(2) = k$
$\ln(3 \times 2) = k$
$\ln(6) = k$
Therefore,$e^k = 6$.

(A) For the logarithm $\log_{10}(A)$ to be defined in $R$,the argument $A$ must be strictly greater than $0$.
Thus,$(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$.
Note that $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$.
So,$(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$.
Since the base $\frac{8}{5} > 1$,the inequality holds for the exponents:
$1 - x^2 > -6(1 + x)$
$1 - x^2 > -6 - 6x$
$x^2 - 6x - 7 < 0$
Factoring the quadratic: $(x - 7)(x + 1) < 0$.
The solution to this inequality is $x \in (-1, 7)$.

(D) For the function $f(x) = \log_2(2^x - 2) + \sqrt{1 - x}$ to be defined,both the logarithmic and square root terms must be real.
$1$. For the square root,we require $1 - x \geq 0$,which implies $x \leq 1$.
$2$. For the logarithm,we require the argument to be positive: $2^x - 2 > 0$,which implies $2^x > 2^1$,so $x > 1$.
Combining these two conditions,we need $x \leq 1$ $AND$ $x > 1$.
Since there is no real number $x$ that satisfies both $x \leq 1$ and $x > 1$ simultaneously,the set of such values is the empty set,denoted by $\phi$.

(C) Given that $x = \log \left( y + \sqrt{y^2 + 1} \right)$.
By the definition of the inverse hyperbolic sine function,we know that $\sinh^{-1}(y) = \log \left( y + \sqrt{y^2 + 1} \right)$.
Comparing the given equation with the definition,we get $\sinh^{-1}(y) = x$.
Taking the hyperbolic sine of both sides,we obtain $y = \sinh x$.

(C) Given that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$.
Note that $\frac{a}{10} + \frac{b}{100} = \frac{1}{10} \left(a + \frac{b}{10}\right)$.
Let $k = a + \frac{b}{10}$. Then the given equations are $k^x = 1000$ and $\left(\frac{k}{10}\right)^y = 1000$.
From $k^x = 1000$,we have $k = 1000^{1/x} = 10^{3/x}$.
From $\left(\frac{k}{10}\right)^y = 1000$,we have $\frac{k}{10} = 1000^{1/y} = 10^{3/y}$.
Thus,$k = 10 \cdot 10^{3/y} = 10^{1 + 3/y}$.
Equating the two expressions for $k$: $10^{3/x} = 10^{1 + 3/y}$.
Comparing the exponents: $\frac{3}{x} = 1 + \frac{3}{y}$.
Rearranging the terms: $\frac{3}{x} - \frac{3}{y} = 1$.
Dividing by $3$: $\frac{1}{x} - \frac{1}{y} = \frac{1}{3}$.

(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$,we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$.
Given $y = \log_9 81$. Since $81 = 9^2$,we have $y = \log_9 9^2 = 2 \log_9 9 = 2(1) = 2$.
Now,we need to evaluate $\sqrt{x - 2\sqrt{y}}$.
Substituting the values of $x$ and $y$,we get $\sqrt{3 - 2\sqrt{2}}$.
We can write $3 - 2\sqrt{2}$ as $(\sqrt{2})^2 + (1)^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$.
Therefore,$\sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.

(A) Given,$a^x = b^y = c^z = d^w = k$ (let).
Taking log with base $a$:
$x = y \log_a b = z \log_a c = w \log_a d$.
Thus,$\frac{1}{y} = \frac{\log_a b}{x}$,$\frac{1}{z} = \frac{\log_a c}{x}$,and $\frac{1}{w} = \frac{\log_a d}{x}$.
Now,$x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right) = x\left(\frac{\log_a b}{x} + \frac{\log_a c}{x} + \frac{\log_a d}{x}\right)$.
$= x \cdot \frac{1}{x} (\log_a b + \log_a c + \log_a d)$.
$= \log_a(bcd)$.

(D) Given the expression: $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$
We can simplify the denominator by factoring out $10^{-4}$:
$10^{-4} a + 10^{-3} b + 10^{-2} c = 10^{-4}(a + 10b + 10^2c)$
Substituting this back into the expression:
$\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4}(a+10 b+10^2 c)}\right)$
$= \log _{10}\left(\frac{1}{10^{-4}}\right)$
$= \log _{10}(10^4)$
$= 4 \log _{10}(10) = 4(1) = 4$

(B) Given the equation: $6^x = 7^{x+4}$
Taking logarithm on both sides: $x \log 6 = (x+4) \log 7$
Using the property $\log(mn) = \log m + \log n$: $x(\log 2 + \log 3) = x \log 7 + 4 \log 7$
Substituting the given values $a, b, c$: $x(a+b) = xc + 4c$
Rearranging the terms to solve for $x$: $x(a+b-c) = 4c$
Therefore: $x = \frac{4c}{a+b-c}$

(A) Given equation: $(x^2 \log _x 27) \cdot \log _9 x = x + 4$
Using the property $\log _a b \cdot \log _b c = \log _a c$,we have $\log _x 27 \cdot \log _9 x = \log _9 27$.
So,the equation becomes $x^2 \cdot \log _9 27 = x + 4$.
Since $\log _9 27 = \log _{3^2} 3^3 = \frac{3}{2} \log _3 3 = \frac{3}{2}$,the equation is $x^2 \cdot \frac{3}{2} = x + 4$.
Multiplying by $2$,we get $3x^2 = 2x + 8$,which simplifies to $3x^2 - 2x - 8 = 0$.
Factoring the quadratic: $3x^2 - 6x + 4x - 8 = 0 \Rightarrow 3x(x - 2) + 4(x - 2) = 0$.
This gives $(3x + 4)(x - 2) = 0$,so $x = 2$ or $x = -\frac{4}{3}$.
Since the base of a logarithm $x$ must be positive and $x \neq 1$,we reject $x = -\frac{4}{3}$.
Therefore,$x = 2$.

(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$.
Using the property $n \log a = \log (a^n)$,we get $\log (x+1)^2 - \log (x^2-1) = \log 2$.
Using the property $\log a - \log b = \log (a/b)$,we get $\log \left( \frac{(x+1)^2}{x^2-1} \right) = \log 2$.
Since $x^2-1 = (x-1)(x+1)$,the equation becomes $\log \left( \frac{(x+1)^2}{(x-1)(x+1)} \right) = \log 2$.
Simplifying the fraction,we have $\log \left( \frac{x+1}{x-1} \right) = \log 2$.
Therefore,$\frac{x+1}{x-1} = 2$.
$x+1 = 2(x-1) \Rightarrow x+1 = 2x-2$.
$x = 3$.
For the logarithmic terms to be defined,we require $x+1 > 0$ and $x^2-1 > 0$,which implies $x > 1$.
Thus,$x = 3$ is the only valid solution.

(A) Taking $\log _{3}$ on both sides,we get:
$(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5 = \log _{3} (3 \sqrt{3}) = \log _{3} (3^{3/2}) = \frac{3}{2}$.
Let $t = \log _{3} x$. Then the equation becomes:
$t^{2} - \frac{9}{2} t + 5 = \frac{3}{2}$.
Multiplying by $2$,we get $2t^{2} - 9t + 10 = 3$,which simplifies to $2t^{2} - 9t + 7 = 0$.
Factoring the quadratic: $(2t - 7)(t - 1) = 0$.
Thus,$t = 1$ or $t = \frac{7}{2}$.
For $t = 1$,$\log _{3} x = 1 \Rightarrow x = 3^{1} = 3$.
For $t = \frac{7}{2}$,$\log _{3} x = \frac{7}{2} \Rightarrow x = 3^{7/2} = 27\sqrt{3}$.
Both roots are real. Therefore,the equation has at least one real root.

(B) Given equation: $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$
Using $\log _{2} 6 = \log _{2} (2 \times 3) = 1 + \log _{2} 3$,we get:
$1 + \log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$
$\log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8) - 1$
$\log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8) - \log _{2} 2$
$\log _{2} 3 + \frac{1}{2x} = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{2} \right)$
$\frac{1}{2x} = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{2 \times 3} \right) = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{6} \right)$
$2^{\frac{1}{2x}} = \frac{2^{\frac{1}{x}} + 8}{6}$
Let $y = 2^{\frac{1}{2x}}$,then $y^2 = 2^{\frac{1}{x}}$.
$6y = y^2 + 8$
$y^2 - 6y + 8 = 0$
$(y - 4)(y - 2) = 0$
$y = 4$ or $y = 2$
If $2^{\frac{1}{2x}} = 4 = 2^2$,then $\frac{1}{2x} = 2 \Rightarrow x = \frac{1}{4}$.
If $2^{\frac{1}{2x}} = 2 = 2^1$,then $\frac{1}{2x} = 1 \Rightarrow x = \frac{1}{2}$.

(C) Given equation: $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$
Rearranging the terms:
$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x \log _{10} 10 + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = \log _{10} 5^{x} - \log _{10} 10^{x} + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = \log _{10}\left(\frac{5^{x} \cdot 6}{10^{x}}\right)$
Taking antilog on both sides:
$1+2^{x} = \frac{5^{x} \cdot 6}{2^{x} \cdot 5^{x}}$
$1+2^{x} = \frac{6}{2^{x}}$
Let $2^{x} = t$. Then $1+t = \frac{6}{t}$
$t + t^{2} = 6$
$t^{2} + t - 6 = 0$
$(t+3)(t-2) = 0$
Since $t = 2^{x} > 0$,we have $t = 2$.
$2^{x} = 2^{1} \Rightarrow x = 1$.

(A) Given equation: $(\log _{5} x)(\log _{x} 3x)(\log _{3x} y) = \log _{x} x^{3}$
Using the change of base formula $\log _{b} a = \frac{\log a}{\log b}$,we get:
$\frac{\log x}{\log 5} \times \frac{\log 3x}{\log x} \times \frac{\log y}{\log 3x} = 3 \log _{x} x$
Simplifying the left side by cancelling common terms:
$\frac{\log y}{\log 5} = 3(1)$
$\log _{5} y = 3$
Converting to exponential form:
$y = 5^{3} = 125$

(C) We have,$\log _{0.2}(x-1) > \log _{0.04}(x+5)$
$\Rightarrow \log _{0.2}(x-1) > \log _{0.2^{2}}(x+5)$
$\Rightarrow \log _{0.2}(x-1) > \frac{1}{2} \log _{0.2}(x+5)$
$\Rightarrow 2 \log _{0.2}(x-1) > \log _{0.2}(x+5)$
$\Rightarrow \log _{0.2}(x-1)^{2} > \log _{0.2}(x+5)$
$\Rightarrow (x-1)^{2} < x+5$
$[\because \log _{a} x > \log _{a} y \Rightarrow x < y, \text{ if } 0 < a < 1]$
$\Rightarrow x^{2}-2x+1 < x+5$
$\Rightarrow x^{2}-3x-4 < 0$
$\Rightarrow (x-4)(x+1) < 0$
$\Rightarrow x \in (-1, 4)$
Also,for the logarithm to be defined,$x-1 > 0 \Rightarrow x > 1$ and $x+5 > 0 \Rightarrow x > -5$.
Combining $x \in (-1, 4)$ and $x > 1$,we get $x \in (1, 4)$.

(C) Given,$\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$
$\therefore \log _{7}(\sqrt{x+7}+\sqrt{x}) = (101)^{0} = 1$
$\Rightarrow \sqrt{x+7}+\sqrt{x} = 7^{1} = 7$
Squaring both sides,we get:
$(\sqrt{x+7}+\sqrt{x})^{2} = 7^{2}$
$x+7+x+2\sqrt{x(x+7)} = 49$
$2x+7+2\sqrt{x^{2}+7x} = 49$
$2\sqrt{x^{2}+7x} = 42-2x$
$\sqrt{x^{2}+7x} = 21-x$
Squaring both sides again:
$x^{2}+7x = (21-x)^{2}$
$x^{2}+7x = 441 - 42x + x^{2}$
$7x = 441 - 42x$
$49x = 441$
$x = \frac{441}{49} = 9$

(C) Let $y = 20^{301}$.
To find the number of digits in $y$,we calculate $\lfloor \log_{10} y \rfloor + 1$.
$\log_{10} y = \log_{10} (20^{301}) = 301 \times \log_{10} (2 \times 10)$.
Using the property $\log(ab) = \log a + \log b$,we get:
$\log_{10} y = 301 \times (\log_{10} 2 + \log_{10} 10) = 301 \times (0.3010 + 1) = 301 \times 1.3010$.
$301 \times 1.3010 = 391.601$.
The number of digits is $\lfloor 391.601 \rfloor + 1 = 391 + 1 = 392$.

(B) Given equation is $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$.
For the logarithm to be defined,we must have $\frac{x+1}{x+5} > 0$ and $x+5 \neq 0$.
This implies $x \in (-\infty, -5) \cup (-1, \infty)$.
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we have $\log_{\sqrt{3}} y = \log_{3^{1/2}} y = 2 \log_3 y$ and $\log_9 y = \log_{3^2} y = \frac{1}{2} \log_3 y$.
The equation becomes $\frac{1}{2} \cdot 2 \log_3 \left(\frac{x+1}{x+5}\right) + \frac{1}{2} \cdot 2 \log_3 |x+5| = 1$.
$\log_3 \left(\frac{x+1}{x+5}\right) + \log_3 |x+5| = 1$.
Since $x+5$ must be positive for the domain,$|x+5| = x+5$.
$\log_3 \left(\frac{x+1}{x+5} \cdot (x+5)\right) = 1$.
$\log_3 (x+1) = 1$.
$x+1 = 3^1 = 3$,so $x = 2$.
Since $x=2$ satisfies the domain condition,there is only $1$ solution.

(C) Given equation: $\log _{2}\left(x^{2}+2 x-1\right)=1$
By the definition of logarithm,$\log _{b}(a) = c \implies a = b^{c}$.
So,$x^{2}+2 x-1 = 2^{1} = 2$.
Rearranging the terms: $x^{2}+2 x-3 = 0$.
Factoring the quadratic equation: $(x+3)(x-1) = 0$.
This gives $x = -3$ or $x = 1$.
Check the domain: For $\log _{2}(f(x))$ to be defined,$f(x) > 0$.
If $x = 1$,$x^{2}+2 x-1 = 1+2-1 = 2 > 0$ (Valid).
If $x = -3$,$x^{2}+2 x-1 = 9-6-1 = 2 > 0$ (Valid).
Both solutions are valid,so there are $2$ solutions.