Question

How do you solve `4^x-2^(x+1)=3`?

Answer 1

`x = log 3/log 2`..

Explanation:

Use `a^(mn)=(a^m)^n=(a^n)^m and a^(m+n)=a^ma^n`

`4^x=(2^2)^x=2^(2x)=(2^x)^2`

The given equation is `u^2-2u-3=0`, where `u = 2^x`
The roots are u = `2^x =3 and -1`. As `2^x>0`, for all `x, -1` is inadmissible..

Now solve `2^x=3`, by equating the logarithms.

`x=log 3/log 2` .