Thread: Solving an exponential equation - Check my method

Hi,

I'm new to the forum, but decided to give it a try now, because I can't seem to get around what I'm doing wrong. I'm revising for my IB finals and can't seem to get this one question down, which is from a past final exam:

So the question is: Solve the equation 4^(x-1)=2^x+8

My method:
4^(x-1) = 2^x+8
4^(x-1) = 2^x+2^3
2^(2(x-1)) = 2^x+2^3
2^(2x-2) = 2^x+2^3 | log
log 2 (2^(2x-2)) = log 2 (2^x) + log 2 (8)
2x-2(log 2 (2)) = x (log 2 (2)) + 3
2x-2 = x + 3
2x-x = 3 + 2
x = 5

But this doesn't check out with the original equation, because it comes to 256 = 40
It's been a while since we did logarithms and exponents, so I might be doing something incorrect. So please correct me where I'm wrong.

Thanks in advance.

Originally Posted by FinnSkies

Hi,

I'm new to the forum, but decided to give it a try now, because I can't seem to get around what I'm doing wrong. I'm revising for my IB finals and can't seem to get this one question down, which is from a past final exam:

So the question is: Solve the equation

My method:
4^(x-1) = 2^x+8
4^(x-1) = 2^x+2^3
2^(2(x-1)) = 2^x+2^3
2^(2x-2) = 2^x+2^3
log 2 (2^(2x-2)) = log 2 (2^x) + log 2 (8) Unfortunately, no. This is the mistake. Taking logs of both sides will lead you to a dead end here. Review this stage carefully, you've not taken logs of the right hand side correctly.

Starting at this stage:

Write it like this:



Let



And solve for Y

Hello, FinnSkies!

Another approach . . .

. . . .
. .
. . .
. . . . . . Correct!

We have: .

I saw that I could turn it into a quadratic if I multiplied by

. .


Factor: .


Then we have: .

. . . . . . . . . . .