Question

If $x,y,z$ are in GP and $a^x=b^y=c^z$, then

• A. ${\log }_{a}c={\log }_{b}a$
• B. ${\log }_{b}a={\log }_{c}b$
• C. ${\log }_{c}b={\log }_{a}c$
• D. None of these

Answer 1

The correct option is B

${\log }_{b}a={\log }_{c}b$

Explanation for the correct answer:

Let $a^x=b^y=c^z=\lambda$ say

Taking logarithm we get

$\log a^x=\log b^y=\log c^z=\log \lambda$

$\Rightarrow$$x\log a=y\log b=z\log c=\log \lambda$ $...\left[\because \log m^n=n\times \log m\right]$

$\Rightarrow$$x=\frac{\log \lambda }{\log a}$ $...\left(i\right)$

$\Rightarrow$$y=\frac{\log \lambda }{\log b}$ $...\left(ii\right)$

$\Rightarrow$ $z=\frac{\log \lambda }{\log c}$ $...\left(iii\right)$

We know that $x,y,z$ are in a geometric progression

$\Rightarrow$$y^2=xz$

From $\left(i\right),\left(ii\right),\left(iii\right)$ we get

$\Rightarrow$ ${\left(\frac{\log \lambda }{\log b}\right)}^2=\frac{\log \lambda }{\log a}\times \frac{\log \lambda }{\log c}$

$\Rightarrow$$\frac{1}{\log b\times \log b}=\frac{1}{\log a\times \log c}$

$\Rightarrow$ $\frac{\log a}{\log b}=\frac{\log b}{\log c}$ $...\left[\because \frac{\log x}{\log y}={\log }_{y}x\right]$

$\Rightarrow$ ${\log }_{b}a={\log }_{c}b$

Hence, option (B) is the correct answer.