The Purplemath Forums

by **danlers** on Tue Aug 23, 2011 12:48 am

This is similar to some other equations which have been solved here, but the "2x" is throwing me, I'm not sure what to do with it.
The equation is 4^x-1 = 3^2x, solve for x. I have the answer, but have not been able to reproduce the steps to get to it.
log4/(log4)-(2log3), then it is completed on a calculator.

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by **stapel_eliz** on Tue Aug 23, 2011 2:16 am

danlers wrote:The equation is 4^x-1 = 3^2x, solve for x.

What you have posted means this:

. . . . . $4^x\, -\, 1\, =\, \left(3^2\right)x$

Was this what you meant? :wink:

by **danlers** on Tue Aug 23, 2011 3:38 am

Oops, no, not quite.
4^(x-1) = 3^(2x)
The exponents are x-1 and 2x.

by **maggiemagnet** on Tue Aug 23, 2011 4:05 pm

danlers wrote:4^(x-1) = 3^(2x)
The exponents are x-1 and 2x.

Start by taking the log of each side. Then use log rules to "expand" the left side and get the variable outside. Then solve for x.

:clap:

by **danlers** on Tue Aug 23, 2011 8:41 pm

Yes, I know how to take the log and expand it, but I am having trouble solving it for x. It seems that everything I do to solve for (x-1) messes up (2x) and vice versa. I'm probably overlooking something obvious.

by **maggiemagnet** on Tue Aug 23, 2011 10:57 pm

danlers wrote:Yes, I know how to take the log and expand it, but I am having trouble solving it for x. It seems that everything I do to solve for (x-1) messes up (2x) and vice versa. I'm probably overlooking something obvious.

Please show your work. Thanks!

:clap:

by **danlers** on Wed Aug 24, 2011 12:09 am

I've tried to solve this so many ways I don't even know what work to show. Let me ask a specific question. If I start by taking the log of both sides, should I do log (base 4), log (base 3), or log (base 10)?

by **maggiemagnet** on Wed Aug 24, 2011 12:52 am

danlers wrote:If I start by taking the log of both sides, should I do log (base 4), log (base 3), or log (base 10)?

Like it showed in the lesson in the link, it doesn't matter.

:clap:

by **danlers** on Wed Aug 24, 2011 5:53 am

Hmm, I see that. Apparently my problem is further on.
Here's one of the ways I've tried it.
4^(x-1) = 3^(2x)
log4^(x-1) = log3^(2x)
x(log4)-1(log4) = 2x(log3)
x(log4)=2x(log3)+(log4)
x/2x=(log3)+(log4)/log4
and if I go a step further I would end up with
x/x=2(log3)+(log4)/log4
and x/x equals one. I always end up with some variation of this, where the x's cancel each other out. I can't get a single x isolated.
Can you point me in the right direction?
Thanks!

by **maggiemagnet** on Wed Aug 24, 2011 11:51 am

It looks like you skipped some steps maybe.

danlers wrote:...x(log4)=2x(log3)+(log4)
x/2x=(log3)+(log4)/log4

How did you go from here:

$x\,\log(4)\, =\, 2x\,\log(3)\, +\, \log(4)$

$x\,\log(4)\, -\, 2x\,\log(3)\, =\, \log(4)$

to here:

$\frac{x}{2x}\, =\, \log(3)\, +\, \frac{\log(4)}{\log(4)}$

:clap:

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