Solving Log Equations (Different Bases)

[MPQC]

Homework Statement

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

Homework Equations

N/A

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?

 

[rock.freak667]

from here

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

the next line should be

lg2+lg3^x=lg7+lg5^x

 

[MPQC]

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

 

[HallsofIvy]

The crucial point is that [itex]2(3^x)[/itex] is NOT [itex](2(3))^x[/itex] so that [itex]ln(2(3^x)[/itex] is NOT x ln(2(3)).

 

[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