# Solve x using logarithms

• Jul 24th 2010, 12:42 AM
DavidM
Solve x using logarithms
I have this problem which I can't seem to work out how to solve. I can get the answer, using Wolfram|Alpha, but I want to know the steps of solving it.

3^(2x) + 3^x = 20

I know x = log(4)/log(3) = 1.2618595...

But how do you work it out? Can someone show me the steps?

The thing that is really confusing me is the plus in the middle, i can't go: log(3^(2x)) + log(3^x) = log(20) because it doesn't work.
What do you put in the middle: log(3^(2x)) ? log(3^x) = log(20) or am i tackling this completely wrong?

No matter how I change/re-arrange the equation, I can't seem to get rid of the plus to put logs into it
• Jul 24th 2010, 03:07 AM
HallsofIvy
Quote:

Originally Posted by DavidM
I have this problem which I can't seem to work out how to solve. I can get the answer, using Wolfram|Alpha, but I want to know the steps of solving it.

3^(2x) + 3^x = 20

I know x = log(4)/log(3) = 1.2618595...

But how do you work it out? Can someone show me the steps?

The thing that is really confusing me is the plus in the middle, i can't go: log(3^(2x)) + log(3^x) = log(20) because it doesn't work.

That's right: log(a+ b) is NOT equal to log(a)+ log(b)

Quote:

What do you put in the middle: log(3^(2x)) ? log(3^x) = log(20) or am i tackling this completely wrong?

No matter how I change/re-arrange the equation, I can't seem to get rid of the plus to put logs into it
Yes, you are "tackling this completely wrong". First recognize that \$\displaystyle 3^{2x}= (3^x)^2\$ so your equation is the same as \$\displaystyle (3^x)^2+
(3^x)= 20\$. If you let \$\displaystyle y= 3^x\$ your equation becomes \$\displaystyle y^2+ y= 20\$. Solve that quadratic equation for y, then use logarithms to solve \$\displaystyle 3^x= y\$.