2*[log3(2x-5) - log32] = log3(2x-3.5) - log32
2*log3$\frac{2^x-5}{2}$ = log3$\frac{2^x-3.5}{2}$
$(\frac{2^x-5}{2})^{2}$ = $\frac{2^x-3.5}{2}$
let $2^{x}$ = a
$(a-5)^{2}$ = 2*(a-3.5)
Solving we get a = 8,4
$2^{x}$ = 8 x = 3
$2^{x}$ = 4 x = 2 Now log3($2^x$-5) = log3(-1) not defined
Answer = b