Question

Sometimes to solve an equation, we may use the identity $a^{{\log }_{a}b}=b,b>0,a>0,a\ne 1$.

Then the number of solutions of the equation $2^{{\log }_6(-4x)}={\log }_{7}2401$ is,

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is B $1$

$2^{{\log }_6(-4x)}={\log }_{7}2401$

above equation is valid when $x<0$
$2^{{\log }_6(-4x)}$

$={\log }_{7}2401={\log }_7{7}^4$

$=4{\log }_{7}7$ ...... $[\because {\log }_{a}a=1]$

$=4=2^2$

Equating the power of 2

$\Rightarrow {\log }_6(-4x)=2$

$\Rightarrow \frac{\log (-4x)}{\log 6}=2$

$\Rightarrow \log (-4x)=2\log 6=\log 6^2$

$\Rightarrow -4x=6^2=36$

$\Rightarrow x=-9$
Ans: B