# Thread: Solving an exponential equation - Check my method

1. ## 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.

2. 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 $\displaystyle 4^{x-1}=2^x+8$

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 2 (2^(2x-2)) = log 2 (2^x) + log 2 (8) Unfortunately, no. This is the mistake. $\displaystyle Log(a+b)\neq Log(a)+Log(b)$ 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: $\displaystyle 2^{2x-2} = 2^x+2^3$

Write it like this:

$\displaystyle \displaystyle\frac{(2^{2x})}{2^2}-2^x-2^3=0$

Let $\displaystyle 2^x=Y$

$\displaystyle \displaystyle\frac{Y^2}{4}-Y-8=0$

And solve for Y

3. Hello, FinnSkies!

Another approach . . .

$\displaystyle \text{Solve: }\:4^{x-1} \:=\:2^x+8$

$\displaystyle \text{My method:}$
. . . . $\displaystyle 4^{x-1} \:=\: 2^x+8$
. . $\displaystyle (2^2)^{x-1} \:=\: 2^x+2^3$
. . .$\displaystyle 2^{2(x-1)} \:=\: 2^x+2^3$
. . . .$\displaystyle 2^{2x-2} \:=\: 2^x+2^3$ . . Correct!

We have: .$\displaystyle 2^{2x-2} - 2^x - 8 \:=\:0$

I saw that I could turn it into a quadratic if I multiplied by $\displaystyle 2^2\!:$

. . $\displaystyle 2^2\!\cdot\!2^{2x-2} - 2^2\!\cdot\!2^x - 2^2\!\cdot\!8 \:=\:0 \quad\Rightarrow\quad 2^{2x} - 4\!\cdot\!2^x - 32 \:=\:0$

Factor: .$\displaystyle (2^x + 4)(2^x - 8) \:=\:0$

Then we have: .$\displaystyle 2^x+4\:=\:0 \quad\Rightarrow\quad 2^x \:=\:-4\quad\text{no real roots}$

. . . . . . . . . . . $\displaystyle 2^x - 8 \:=\:0 \quad\Rightarrow\quad 2^x \:=\:8 \quad\Rightarrow\quad \boxed{x \:=\:3}$