Question

If $lo{g}^2,log(2^x-1)$ and $log(2^x+3)$ are in A.P., write the value of x.

Answer 1

Since log 2, $log(2^x-1)$ and $log(2^x+3)$ are in A.P., therefore

$log(2^x-1)-log2=log(2^x+3)-log(2^x-1)$

$2log(2^x-1)=log(2^x+3)+log2$

$(2x-1{)}^2=2(2^x+3)$

$2^{2x}-{2.2}^x+1={2.2}^x+6$

$2^{2x}-{4.2}^x-5=0$

$2^{2x}+2^x-{5.2}^x-5=0$

$2^x(2^x+1)-5(2^x+1)=0$

$(2^x+1)(2^x-5)=0$

2x - 5 = 0

2x = 5

$xlog2=log5$

$x=\frac{log5}{log2}$

$=lo{g}_{2}5$