Question

Find \(x\).

\[2^{2x} = 8^{x-1}\]
\[8^{x-1}\Rightarrow (2^3)^{x-1}\]
\[2^{2x} = (2^3)^{x-1}\]
What do I do from here on?

Answer 1

Apply the laws of exponents on the right-hand side \[\Large (a^n)^m = a^{mn}\]

Answer 2

Ah, so I basically distribute?

\[2^{2x} = (2^3)^{x-1}\]
\[2^{2x} = 2^{3x-3}\]
So from there:
\[2x = 3x - 3\]
\[-x = -3\]
\[x = 3\]

I have a question though, what if the bases were different?

Answer 3

Like?
(The bases WERE different, by the way)

Answer 4

But strictly speaking, if you have something like this (I'm going to be very general here)
\[\Large p^{f(x)}= q^{g(x)}\]

Answer 5

Such as:
\[2^{2x} = 4^{3x-3}\]
Would that effect my answer in finding \(x\)?

Answer 6

Well yes, but this case is actually rather mild, since you know fully well that
\[\Large 2^2 = 4\implies=4^{3x-3}= 2^{2(3x-3)}\]

Answer 7

What if it wasn't mild such as going to 5 or some other odd number?

Answer 8

In general, when given
\[\Large p^{f(x)}= q^{g(x)}\]

The way to do it is to realise that
\[\Huge q = p^{\log_p (q)}\]
So that you get
\[\Huge p^{f(x)}= p^{[g(x)]\log_p(q)}\]

And finally
\[\Large f(x) = g(x)\log_p(q) \]

Answer 9

If we apply this to the first question, we, in fact, get
\[\Large p = 2 \\\Large q=8\\\Large f(x) = 2x\\\Large g(x) = x-1\]

Following that rule, we get
\[\Large f(x) = g(x) \log_p(q)\]\[\Large 2x= (x-1)\log_2(8)\]\[\Large 2x = (x-1)\cdot 3 \]
etc

Answer 10

But this is 'mild' since 8 is a power of 2.

The uglier solutions arise when q is not a power of p (or vice versa)

Answer 11

I see, thank you.