Question

Consider the system of the equations: $(ax{)}^{loga}=(by{)}^{logb};b^{logx}=a^{logy}$ where $a>0,b>0$ and $a\ne b,ab\ne 1$
x and y will be:

• A. $x=\frac{1}{a}$ and $y=\frac{1}{b}$
• B. $x=\frac{-1}{a}$ and $y=\frac{-1}{b}$
• C. $x=\frac{1}{a+b}$ and $y=\frac{1}{a-b}$
• D. $x=\frac{1}{a-b}$ and $y=\frac{1}{a+b}$

Answer 1

The correct option is A $x=\frac{1}{a}$ and $y=\frac{1}{b}$

Given, $(ax{)}^{loga}=(by{)}^{logb}--->1,b^{logx}=a^{logy}--->2$
Applying logarithm on both sides of the above two equations,
$log(ax{)}^{loga}=log(by{)}^{logb}$
$\Rightarrow loga\ast (loga+logx)=logb\ast (logb+logy)$--->A
Similarly appling logarith on equation 2,
$log{b}^{logx}=log{a}^{logy}$
$\Rightarrow logblogx=loga\ast logy$---->B
On substituting logy from equation B into equation A,
$\Rightarrow loga.(loga+logx)=logb\ast (logb+\frac{logb.logx}{loga})$
$\Rightarrow logx\ast (loga-\frac{logb}{loga})=(logb{)}^2-(loga{)}^2$
$\Rightarrow -\frac{logx}{loga}=1$
$\Rightarrow loga+logx=0$
$\Rightarrow logax=0$
$\Rightarrow ax=1$
$\Rightarrow x=\frac{1}{a}$--->C
on substituting C in B,
$\Rightarrow logb\ast log\left(\frac{1}{a}\right)=loga\ast logy$
$\Rightarrow logb\ast (-loga)=loga\ast logy$
$\Rightarrow -logb=logy$
$\Rightarrow logby=0$
$\Rightarrow y=\frac{1}{b}$
Option A is correct.