- Fri Aug 26, 2011 2:03 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

Thanks, I got it figured out.

- Wed Aug 24, 2011 4:01 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

By doing it too late at night...

The fact is I don't really know what to do at that point. That's the problem.

The fact is I don't really know what to do at that point. That's the problem.

- Wed Aug 24, 2011 5:53 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

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 ...

- Wed Aug 24, 2011 12:09 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

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)?

- Tue Aug 23, 2011 8:41 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

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.

- Tue Aug 23, 2011 3:38 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

Oops, no, not quite.

4^(x-1) = 3^(2x)

The exponents are x-1 and 2x.

4^(x-1) = 3^(2x)

The exponents are x-1 and 2x.

- Tue Aug 23, 2011 12:48 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithmic equation 4^x-1 = 3^2x?
- Replies:
**12** - Views:
**24447**

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 ...