Question

If $\log \left(a\right)=\frac{1}{2}\log \left(b\right)=\frac{1}{5}\log \left(c\right)$, then $a^4{b}^3{c}^{-2}=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1: Solve for the values of $a,b,c$

Let $\log \left(a\right)=\frac{1}{2}\log \left(b\right)=\frac{1}{5}\log \left(c\right)=k$

Assume base to be $e$ since base is not given

$\Rightarrow$ $\log a=k$

$\Rightarrow$ $a=e^k$ $...\left(i\right)$ $\left[\because {\log }_y\left(x\right)=z\Rightarrow x=y^z\right]$

Similarly,

$\frac{1}{2}\log b=k$

$\Rightarrow$ $\log b=2k$

$\Rightarrow$ $b=e^{2k}$ $...\left(ii\right)$

$\Rightarrow$ $\frac{1}{5}\log c=k$

$\Rightarrow$ $\log c=5k$

$\Rightarrow$ $c=e^{5k}$ $...\left(iii\right)$

Step 2: Solve for the required value

$a^4{b}^3{c}^{-2}=\frac{a^4{b}^3}{c^2}$

Substitute all values from $\left(i\right),\left(ii\right),\left(iii\right)$

$\Rightarrow$ $a^4{b}^3{c}^{-2}=\frac{{{\left(e^k\right)}^4\left(e^{2k}\right)}^3}{{\left(e^{5k}\right)}^2}$

$\Rightarrow$ $a^4{b}^3{c}^{-2}=\frac{e^{4k}\times e^{6k}}{e^{10k}}$

$\Rightarrow$ $a^4{b}^3{c}^{-2}=\frac{e^{10k}}{e^{10k}}$ $\left[\because a^m\times a^n=a^{m+n}\right]$

$\Rightarrow$ $a^4{b}^3{c}^{-2}=1$

Hence the value of $a^4{b}^3{c}^{-2}$ is $1$

Hence, option(A) i.e. $1$ is the correct answer.