Question

Solve the equation ?

`4^2` + `2^(2x + 3)` = 40 + `4^(x+1)`

Answer 1

Real valued solution:

`x = (ln 6)/(2 ln 2)`

All solutions:

`x = (ln 6 + 2npii)/(2 ln 2)" "n in ZZ`

Explanation:

Given:

`4^2+2^(2x+3) = 40+4^(x+1)`

Note that `4=2^2`, so:

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

So we have:

`16+2^(2x+3) = 40+2^(2x+2)`

Subtract `16+2^(2x+2)` from both sides to get:

`2^(2x+3)-2^(2x+2) = 24`

That is:

`(2^3-2^2)2^(2x) = 24`

Dividing both sides by `2^3-2^2 = 2^2 = 4`, we get:

`2^(2x) = 6`

Taking natural logs of both sides:

`2x ln 2 = ln 6`

or for all solutions, including complex ones:

`2x ln 2 = ln 6 + 2npii" "n in ZZ`

Dividing both sides by `2 ln 2`, we get:

`x = (ln 6 + 2npii)/(2 ln 2)" "n in ZZ`