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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

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