logarithm Questions in English

(B) We are given $\log _{10} 2986 = 3.4751$.
We need to find $\log _{10} 0.02986$.
$\log _{10} 0.02986 = \log _{10} \left( \frac{2986}{10^5} \right)$
Using the property $\log \left( \frac{a}{b} \right) = \log a - \log b$,we get:
$= \log _{10} 2986 - \log _{10} 10^5$
$= 3.4751 - 5$
$= -1.5249$
In characteristic-mantissa form,$-1.5249 = -2 + 0.4751 = \overline{2}.4751$.

(A) Given the equation: $\log (2 a-3 b)=\log a-\log b$
Using the logarithmic property $\log x - \log y = \log (x/y)$,we get:
$\log (2 a-3 b)=\log (a/b)$
By taking the antilog on both sides,we have:
$2 a-3 b = a/b$
Multiply both sides by $b$:
$2 a b - 3 b^{2} = a$
Rearrange the terms to isolate $a$:
$2 a b - a = 3 b^{2}$
$a(2 b-1) = 3 b^{2}$
Therefore,$a = \frac{3 b^{2}}{2 b-1}$

(C) Given equation: $\log (x-y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0$
Rearranging the terms: $\log (x-y) = \log 5 + \frac{1}{2} (\log x + \log y)$
Using logarithm properties $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:
$\log (x-y) = \log 5 + \log (xy)^{1/2}$
$\log (x-y) = \log (5 \sqrt{xy})$
Taking antilog on both sides: $x-y = 5 \sqrt{xy}$
Squaring both sides: $(x-y)^2 = (5 \sqrt{xy})^2$
$x^2 - 2xy + y^2 = 25xy$
$x^2 + y^2 = 27xy$
Dividing both sides by $xy$: $\frac{x^2}{xy} + \frac{y^2}{xy} = \frac{27xy}{xy}$
$\frac{x}{y} + \frac{y}{x} = 27$

(B) Given that $\frac{\log x}{3} = \frac{\log y}{4} = \frac{\log z}{5} = k$.
From this,we can express the logarithms as:
$\log x = 3k$,$\log y = 4k$,and $\log z = 5k$.
We want to find the value of $zx$. Using the properties of logarithms,we have:
$\log(zx) = \log z + \log x$.
Substituting the values in terms of $k$:
$\log(zx) = 5k + 3k = 8k$.
Since $\log y = 4k$,we can write $8k = 2(4k) = 2 \log y$.
Using the power rule of logarithms,$2 \log y = \log(y^2)$.
Therefore,$\log(zx) = \log(y^2)$,which implies $zx = y^2$.

(A) Given the equation: $3+\log _{5} x=2 \log _{25} y$
We know that $\log _{25} y = \frac{\log _{5} y}{\log _{5} 25} = \frac{\log _{5} y}{2}$.
Substituting this into the equation: $3+\log _{5} x = 2 \cdot \frac{\log _{5} y}{2}$.
This simplifies to: $3+\log _{5} x = \log _{5} y$.
We can write $3$ as $\log _{5} 125$,so: $\log _{5} 125 + \log _{5} x = \log _{5} y$.
Using the property $\log a + \log b = \log(ab)$,we get: $\log _{5}(125x) = \log _{5} y$.
By comparing both sides: $125x = y$,which implies $x = \frac{y}{125}$.

(D) Let $\frac{\log _{2} a}{2}=\frac{\log _{3} b}{3}=\frac{\log _{4} c}{4}=k$.
From this,we get $\log _{2} a = 2k \Rightarrow a = 2^{2k}$,$\log _{3} b = 3k \Rightarrow b = 3^{3k}$,and $\log _{4} c = 4k \Rightarrow c = 4^{4k}$.
Substitute these into the given equation $a^{1/2} \cdot b^{1/3} \cdot c^{1/4} = 24$:
$(2^{2k})^{1/2} \cdot (3^{3k})^{1/3} \cdot (4^{4k})^{1/4} = 24$
$2^k \cdot 3^k \cdot 4^k = 24$
$(2 \cdot 3 \cdot 4)^k = 24$
$24^k = 24^1 \Rightarrow k = 1$.
Now,substitute $k=1$ back into the expressions for $a, b,$ and $c$:
$a = 2^{2(1)} = 4$
$b = 3^{3(1)} = 27$
$c = 4^{4(1)} = 256$.
Thus,the correct option is $c=256$.

(C) Given $\frac{\log _{2} x}{3}=\frac{\log _{2} y}{4}=\frac{\log _{2} z}{5 k} = \lambda$ (let).
Then $\log _{2} x = 3\lambda$,$\log _{2} y = 4\lambda$,and $\log _{2} z = 5k\lambda$.
Given $\frac{z}{x^{3} y^{4}}=1$,taking $\log _{2}$ on both sides:
$\log _{2} z - \log _{2} (x^{3} y^{4}) = \log _{2} 1$
$\log _{2} z - 3 \log _{2} x - 4 \log _{2} y = 0$.
Substituting the values of $\log _{2} x, \log _{2} y, \log _{2} z$:
$5k\lambda - 3(3\lambda) - 4(4\lambda) = 0$
$5k\lambda - 9\lambda - 16\lambda = 0$
$5k\lambda - 25\lambda = 0$
Since $\lambda \neq 0$,we have $5k - 25 = 0$,which gives $k = 5$.

(A) Given expression: $E = \frac{3 + \log_{10} 343}{2 + \frac{1}{2} \log_{10} \left(\frac{49}{4}\right) + \frac{1}{3} \log_{10} \left(\frac{1}{125}\right)}$
Numerator: $3 + \log_{10} 7^3 = 3 + 3 \log_{10} 7 = 3(1 + \log_{10} 7) = 3(\log_{10} 10 + \log_{10} 7) = 3 \log_{10} 70$.
Denominator: $2 + \log_{10} \left(\frac{49}{4}\right)^{1/2} + \log_{10} \left(\frac{1}{125}\right)^{1/3} = 2 + \log_{10} \left(\frac{7}{2}\right) + \log_{10} \left(\frac{1}{5}\right)$.
Using $\log a + \log b = \log(ab)$,the denominator becomes: $2 + \log_{10} \left(\frac{7}{2} \times \frac{1}{5}\right) = 2 + \log_{10} \left(\frac{7}{10}\right)$.
Since $2 = \log_{10} 100$,the denominator is: $\log_{10} 100 + \log_{10} 0.7 = \log_{10} (100 \times 0.7) = \log_{10} 70$.
Thus,$E = \frac{3 \log_{10} 70}{\log_{10} 70} = 3$.

(C) Let each ratio be equal to a constant $k$.
Then,$\log x = k(a^2 + ab + b^2)$,$\log y = k(b^2 + bc + c^2)$,and $\log z = k(c^2 + ca + a^2)$.
Multiplying the first equation by $(a-b)$,we get:
$(a-b) \log x = k(a-b)(a^2 + ab + b^2) = k(a^3 - b^3)$.
Thus,$\log x^{a-b} = k(a^3 - b^3)$.
Similarly,$(b-c) \log y = k(b^3 - c^3)$ and $(c-a) \log z = k(c^3 - a^3)$.
Now,let $E = x^{a-b} \cdot y^{b-c} \cdot z^{c-a}$.
Taking the logarithm on both sides:
$\log E = \log x^{a-b} + \log y^{b-c} + \log z^{c-a}$
$\log E = k(a^3 - b^3) + k(b^3 - c^3) + k(c^3 - a^3)$
$\log E = k(a^3 - b^3 + b^3 - c^3 + c^3 - a^3) = k(0) = 0$.
Since $\log E = 0$,we have $E = 10^0 = 1$ (assuming common log) or $E = e^0 = 1$ (assuming natural log).
Therefore,$x^{a-b} \cdot y^{b-c} \cdot z^{c-a} = 1$.

(C) Given $3^{x-2} = 5$.
Taking $\log_{10}$ on both sides:
$(x-2) \log_{10} 3 = \log_{10} 5$
Since $\log_{10} 5 = \log_{10} (10/2) = \log_{10} 10 - \log_{10} 2 = 1 - 0.30103 = 0.69897$.
So,$(x-2) (0.4771) = 0.69897$.
$x-2 = \frac{0.69897}{0.4771} = 1.465038...$
$x = 2 + 1.465038 = 3.465038$.
Calculating the fraction: $3 \frac{22187}{47710} = 3 + \frac{22187}{47710} \approx 3 + 0.465038 = 3.465038$.
Thus,$x = 3 \frac{22187}{47710}$.

(D) Let $x = (648)^{5}$.
Taking logarithm on both sides,we get $\log x = 5 \log (648)$.
Since $648 = 2^3 \times 3^4$,we have $\log x = 5 \log (2^3 \times 3^4)$.
Using the properties of logarithms,$\log x = 5 (3 \log 2 + 4 \log 3)$.
Substituting the given values,$\log x = 5 (3 \times 0.30103 + 4 \times 0.4771)$.
$\log x = 5 (0.90309 + 1.9084) = 5 (2.81149) = 14.05745$.
The number of digits in a number $x$ is given by $\lfloor \log_{10} x \rfloor + 1$.
Here,$\lfloor 14.05745 \rfloor + 1 = 14 + 1 = 15$.
Therefore,the number of digits in $(648)^{5}$ is $15$.

(C) Let $\frac{\log x}{1} = \frac{\log y}{2} = \frac{\log z}{5} = k$.
Then,$\log x = k$,$\log y = 2k$,and $\log z = 5k$.
We need to find the value of $x^{4} \cdot y^{3} \cdot z^{-2}$.
Taking the logarithm of the expression,we get:
$\log(x^{4} \cdot y^{3} \cdot z^{-2}) = 4 \log x + 3 \log y - 2 \log z$.
Substituting the values of $\log x, \log y,$ and $\log z$ in terms of $k$:
$= 4(k) + 3(2k) - 2(5k)$
$= 4k + 6k - 10k$
$= 10k - 10k = 0$.
Since $\log(x^{4} \cdot y^{3} \cdot z^{-2}) = 0$,it follows that $x^{4} \cdot y^{3} \cdot z^{-2} = 10^{0} = 1$.

(C) Given expression: $\frac{\log \sqrt{27}+\log \sqrt{1000}+\log 8}{\log 120}$
Step $1$: Simplify the numerator terms.
$\log \sqrt{27} = \log (3^3)^{1/2} = \frac{3}{2} \log 3$
$\log \sqrt{1000} = \log (10^3)^{1/2} = \frac{3}{2} \log 10$
$\log 8 = \log 2^3 = 3 \log 2 = \frac{3}{2} \log 2^2 = \frac{3}{2} \log 4$
Step $2$: Substitute these into the numerator.
Numerator $= \frac{3}{2} \log 3 + \frac{3}{2} \log 10 + \frac{3}{2} \log 4 = \frac{3}{2} (\log 3 + \log 10 + \log 4)$
Step $3$: Use the property $\log a + \log b + \log c = \log (a \cdot b \cdot c)$.
Numerator $= \frac{3}{2} \log (3 \cdot 10 \cdot 4) = \frac{3}{2} \log 120$
Step $4$: Divide by the denominator.
$\frac{\frac{3}{2} \log 120}{\log 120} = \frac{3}{2}$

(B) Given the equation $y = \frac{10^{\log_{10} x}}{x^2}$.
Using the logarithmic property $10^{\log_{10} x} = x$,the equation simplifies to $y = \frac{x}{x^2}$.
This further simplifies to $y = \frac{1}{x}$,which implies $x = \frac{1}{y} = y^{-1}$.
Given that $x = y^a$,by comparing the exponents,we get $a = -1$.

(B) Given $x = \log_{4/3}(1/2)$. Since the base $4/3 > 1$ and the argument $1/2 < 1$,the value of $x$ is negative $(x < 0)$.
Given $y = \log_{1/2}(1/3)$. Using the property $\log_{a^n} b^m = \frac{m}{n} \log_a b$,we have $y = \log_{2^{-1}} (3^{-1}) = \frac{-1}{-1} \log_2 3 = \log_2 3$. Since the base $2 > 1$ and the argument $3 > 1$,the value of $y$ is positive $(y > 0)$.
Comparing the two,since $x < 0$ and $y > 0$,it follows that $x < y$.