Question

@whpalmer4
solve 3^(2x)=108

Answer 1

The left side can be simplified to 3^2 * 3^x, which yields 9*3^x.
We divide both sides by nine to get:
3^x = 9
Now we have a simple equation we can solve using logarithms. Do you know how to solve this?

Answer 2

\[\log_{3} 108=2x\]

Answer 3

\[3^{2x} = 108\]\[\ln 3^{2x} = \ln 108\]\[2x\ln 3 = \ln 108\]\[x = \frac{\ln 108}{2\ln 3}\approx 2.13\]

Answer 4

Okay, or do it the inefficient way.
I am sad now.

Answer 5

I'm sad that you think 108/9 = 9!

Answer 6

wait is in ln or log?

Answer 7

I'm really good at math, I promise. But 3^x = 12 is still a much easier equation to solve.

Answer 8

Arithmetic gets to me sometimes. :C
Anyway, ln is the natural logarithm. Log is used typically in reference to the common logarithm unless otherwise specified.

Answer 9

i came up with 2.5719 ... not 2.13

Answer 10

@murdocjax check your answer.

Answer 11

Well, I wrote out steps for the benefit of the chap who asked me for help reviewing for his test. Normally, I would just write down \[x = \frac{\ln 108}{2 \ln 3}\]without any need to factor out 3^2 and divide both sides. Being able to look at the problem and write the answer immediately makes me think my method is satisfactory, especially when it doesn't cause me to do things that sometimes trip me up, but there are certainly many ways to solve such problems, and everyone should find the way that works best for their needs.

Answer 12

\[3^{2*2.5719} \approx 284.588\]

Answer 13

yes the long way helps to understand the problem

Answer 14

wait now i'm confused. what's the final answer?

Answer 15

Also, \[3^{2x} \ne 9*3^x\]
\[3^{2x} = 3^x*3^x\]

Answer 16

(well, okay, \(3^{2x} = 9*3^x\) if \(x=2\))

@murdocjax my original answer is correct.

Answer 17

I don't know, man. I have a really weird intuition of exponents, so I break things apart a lot.

Answer 18

the 2.13

Answer 19

key word might be "break" :-)

Here's a plot of \(3^{2x}\) and \(9*3^x\) which I think conclusively demonstrates that they are not equal for just about every value of interest here.

Answer 20

@murdocjax you can do the problem with logs of any base, and still get the same answer.

For example, if we do it with \(\log_{10}\):

\[x = \frac{\log_{10} 108}{2\log_{10}3} = \frac{2.03342}{2*0.477121} = 2.13093\]

Do it with \(\log_3\):
\[x = \frac{\log_3 108}{2\log_3 3} = \frac{4.26186}{2*1} = 2.13093\]

Do it with \(\ln\):
\[x = \frac{\ln 108}{2 \ln 3} = \frac{ 4.68213}{2*1.09861} = 2.13093\]

Answer 21

@whpalmer, I think you may want to check your order of operations for those graphs.

Answer 22

I think I meant 9^x. Yeah, I meant that.
I'm messing up a lot. Oh well.

Answer 23

uh huh. http://www.wolframalpha.com/input/?i=3%5E%282x%29%2C+9*3%5Ex+from+x%3D1+to+5

Look, you are asserting that \[3^2*3^x = 3^{2x}\]

Let's try out some values.
\[x=0\]\[3^2*3^0 = 3^{2*0}\]\[9*1 = 1\]Uh, no.

\[x=1\]
\[3^2*3^1 = 3^{2*1}\]\[9*3 = 3^2\]\[27=9\]Uh, no.

Answer 24

"I think I meant 9^x. Yeah, I meant that.
I'm messing up a lot. Oh well."

Answer 25

And Wolfram confirms me on that one.
http://www.wolframalpha.com/input/?i=3%5E%282*x%29

Answer 26

Yes, \[((x^m)^n) = x^{mn}\]
By now the chap has had time to memorize log 2 and log 3 in a number of bases, which is all he would need to evaluate that result as a decimal. \[108 = 2^2*3^3\]\[\log 108 = 2\log 2 + 3\log 3\]