Question

How do I solve 64^(x) = 16^(x-1)?

Answer 1

rewrite them as powers of 4\[(4^3)^x = (4^2)^{x - 1}\]

you can use the power of a power law to simplify them

\[(x^a)^b = x^{a \times b}\]

so once you have simplifed the powers, equate them and solve for x.

Answer 2

So then it's x=-1, right?

Answer 3

nope...

Answer 4

oh, i forgot to multiply the 2 and -1, so x=-2, right?

Answer 5

correct.

Answer 6

Okay, thanks!