# Solving Log Equations (Different Bases)

1. Jan 21, 2010

### MPQC

1. The problem statement, all variables and given/known data

Only the equation: 2(3^x) = 7(5^x)

2. Relevant equations

N/A

3. The attempt at a solution

Okay. I've been fine learning the loglaws, until I hit this question. If they were the same base (or at least near the same, ex: 2 & 8) then I could simply change them into the same base, and proceed. But I can't do that for this question - neither can I just throw "log" it all. (Though that's my solution for now - I'm really not sure how I should proceed.)

So here's what I've tried:

2(3^x) = 7(5^x)
log (2(3^x)) = log (7(5^x))
x log (2(3)) = x log (7(5))
x log (6) = x log (35)
x = x ( log (35) / log (6) )
x = 1.98x

Well that doesn't work out too well.. I've tried googling the answer - but all I get is examples with the same base, or a base that I can change to make them the same. So how could I solve this?

2. Jan 21, 2010

### rock.freak667

from here

log (2(3^x)) = log (7(5^x))

the next line should be

lg2+lg3x=lg7+lg5x

3. Jan 21, 2010

### MPQC

Thanks so much. I had learned to put them together, but the thought of splitting them back up never occurred.

4. Jan 22, 2010

### HallsofIvy

Staff Emeritus
The crucial point is that $2(3^x)$ is NOT $(2(3))^x$ so that $ln(2(3^x)$ is NOT x ln(2(3)).

5. Jan 16, 2012

### violetcat

Here's an alternate solution:
2(3^x)=7(5^x)
2/7=(5^x)/(3^x) Divide both sides by 7 and by 3^x
2/7=(5/3)^x
log(2/7)=log((5/3)^x)
log(2/7)=x log(5/3)
log(2/7)/log(5/3)=x

Last edited: Jan 16, 2012