Question

How to solve for x? : 8*2^2x+4*2^x+1=1+2^x

Answer 1

`x` cannot be solved.

Explanation:

`8*2^(2x)+4*2^x+1=1+2^x`

Subtract `1` from both sides,

`8*2^(2x)+4*2^x=2^x`

Subtract `2^x` from both sides,

`8*2^(2x)+3*2^x=0`

Apply the power of a power law, where `(a^n)^m=a^(nm)`,

`8*(2^x)^2+3*2^x=0`

Let `2^x` be `u`,

`8u^2+3u=0`

Factor,

`u(8u+3)=0`

Solve,

`u=0 or -3/8`

When `u=0`,

`2^x=0` ( No solution as `2^x` cannot be `0` )

When `u=-3/8`,

`2^x=-3/8` ( No solution as `2^x` cannot be negative )

Hence, `x` cannot be solved.