# Thread: Solving for x with logs

1. ## Solving for x with logs

Problem: 2^2x=k

Maybe someone can tell me what to look for. I am going to fiddle with it a little more.

2. ## Re: Solving for x with logs

Originally Posted by benny92000
Problem: 2^2x=k

Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
Either

\displaystyle \begin{align*} 2^{2x} &= k \\ 2x &= \log_2{k} \\ x &= \frac{1}{2}\log_2{k} \end{align*}

or

\displaystyle \begin{align*} 2^{2x} &= k \\ \ln{\left(2^{2x}\right)} &= \ln{k} \\ 2x\ln{2} &= \ln{k} \\ 2x &= \frac{\ln{k}}{\ln{2}} \\ x &= \frac{\ln{k}}{2\ln{2}} \end{align*}

3. ## Re: Solving for x with logs

Originally Posted by benny92000
Problem: 2^2x=k

Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
$2^{2x} = k$

$\log(2^{2x}) = \log{k}$

$2x\log{2} = \log{k}$

$x = \frac{\log{k}}{2\log{2}}$

4. ## Re: Solving for x with logs

Can I also start by taking the log2 of both sides?

5. ## Re: Solving for x with logs

Originally Posted by benny92000
Can I also start by taking the log2 of both sides?
You obviously didn't read my post properly, as that was the first method I showed...

6. ## Re: Solving for x with logs

Originally Posted by benny92000
Problem: 2^2x=k

Maybe someone can tell me what to look for. I am going to fiddle with it a little more.

Start by taking the natural log of both sides of the equation.

Let ln = natural log

2^(2x) = k

ln[2^(2x)] = ln(k)

When we take the natural log of both sides, exponents come down infront of the ln symbol.

So, 2x comes down and it now looks like this:

2xln(2) = ln(k)

At this point, solve the equation for x, dividing both sides by 2(ln2).

x = ln(k)/2(ln2)

7. ## Re: Solving for x with logs

Originally Posted by benny92000
Can I also start by taking the log2 of both sides?
you can use any log base you want.

I believe this thread has been beat to death.