Question

The largest integer satisfying the equation $|4+{\log }_{1/7}x|=2+|2+{\log }_{1/7}x|$ is

Answer 1

$|4+{\log }_{1/7}x|=2+|2+{\log }_{1/7}x|$

Clearly $x>0$.

Put ${\log }_{1/7}x=t$ to obtain

$|4+t|=2+|2+t|$

Case 1:
If $t<-4$
$-4-t=2-2-t$
$\Rightarrow -4=0$
This is always false.
Case 2:
If $-4\le t\le -2$
$t+4=2-2-t$
$t=-2$
Case 3:
If $t>2$
$4+t=2+2+t$
$\Rightarrow 4=4$
From Case 1, Case 2, and Case 3, we get
$t=-2={\log }_{1/7}x=-{\log }_{7}x[\because {\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow x=7^2=49$
Ans: 49