Question

4^k=x^2+8x
how do you solve in terms of k please.

Answer 1

i guess you need to take the

Answer 2

http://freemathhelp.com/

Answer 3

\[k=\log_4(x^2+8x)\]

Answer 4

that's what i had at first but i wasn't sure if you could simplify more

Answer 5

and sorry, it has to be in terms of x

Answer 6

well you could write this


\[k=\log_4(x(x+8))=\log_4(x)+\log_4(x+8)\] but that really doesn't get you any closer

Answer 7

well that is completely different. write
\[x^2+8x-4^k=0\] and then solve via the quadratic formula

Answer 8

it does not matter that there is a 4^k with the quadratic formula?

Answer 9

In the quadratic formula i put 4 as c but i'm not getting the correct answer