Question

How do you solve `4^(log_4(x+8))=4^2`?

Answer 1

Soln. is `x=8.`

Explanation:

Given that, `4^(log_4^(x+8))`=4^2#

Since the bases are same, the indices must be equal, and hence,

`log_4^(x+8)=2`
`rArr x+8=4^2=16`

`rArr x=8`

Verification holds good, so the Soln. is `x=8.`

Answer 2

see below

lots of ways to do this

Explanation:

`4^(log_4(x+8))=4^2`

you can see immediately that `x+8 = 4^2 implies x = 8`

this is because: `a ^ {ln_a Q} = Q`

or you can plod through mechanically. So, just equating the exponents....

`log_color{red}{4}(x+8)=2 qquad star`

`(x+8)=color{red}{4}^2` same result

or changing the base on the LHS of `star` to 2

`log_color{red}{4}(x+8) = (log_2(x+8))/(log_2 4)`

so

`(log_2(x+8))/(log_2 4)=2`

`implies log_2(x+8)=2* log_2 4`

`implies log_2(x+8)=2* 2 = 4`

`implies (x+8)=2^4 = 16 implies x = 8`