Question

Given that 2^x = 7, find the value of 4^x-1.

Answer 1

$4=2^2\\\\4^x=(2^2)^x=(2^x)^2=7^2=49\\\\4^x-1=49-1=48$

$If\ is\ 4^{x-1},\ then\\\\4^{x-1}=\dfrac{4^x}{4}=\dfrac{49}{4}=12\frac{1}{4}$

$Used:\\\dfrac{a^n}{a^m}=a^{n-m}$