Question

How do you solve `8^(2x-4)=16^(2x+4)`?

Answer 1

I got: `x=-14`

Explanation:

Let us take the log in base `2` of both sides and use a property of logs to move the exponents in front of the logs. You should get:
`log_2(8)^(2x-4)=log_2(16)^(2x+4)`
`(2x-4)log_2(8)=(2x+4)log_2(16)`
we now use the definition of log to solve the two logs and get:
`(2x-4)*3=(2x+4)*4`
rearrange and solve for `x`:
`6x-12=8x+16`
`2x=-28`
`x=-28/2=-14`

Answer 2

`x = -14`

Explanation:

Some exponential equations need to be solved using logs, others can be solved by making the bases or the indices the same.

This is an example of the latter.

Both 8 and 16 are powers of 2, so we can write this equation in terms of a common base on each side.

`color(red)(8 = 2^3) and color(blue)(16 = 2^4)color(red)(`

`color(red)(8)^(2x-4) = color(blue)(16)^(2x+4)`

`color(red)((2^3))^(2x-4) = color(blue)((2^4))^(2x+4)" " larr` multiply the indices.

`2^(6x-12) = 2^(8x+16)" "larr` if `x^m = x^n hArr m = n`

`:. 6x-12 = 8x+16`

`-12-16 = 8x-6x`

`-28 = 2x`

`-14 = x`