## Solving for x in 10^(3x-8) = 2^(5-x)

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Aqua Dragon
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### Solving for x in 10^(3x-8) = 2^(5-x)

The title pretty much states the problem. I'm trying to solve for x in 10^(3x-8) = 2^(5-x). The problem is that once I get it down to 3x-8 = 5 log 2 - x log 2, I don't know what to do. I continued trying to simplify it, and ended up with just having the other side log'ified, which was right back at square 1. Any help on this would be appreciated.

stapel_eliz
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I'm trying to solve for x in 10^(3x-8) = 2^(5-x). The problem is that once I get it down to 3x-8 = 5 log 2 - x log 2, I don't know what to do.
Do the same as you'd do for solving the linear equation if it were of the form "3x - 8 = 2 - 2x".

. . . . .$3x\, -\, 8\, =\, 5\log(2)\, -\, x\log(2)$

. . . . .$3x\, +\, x\log(2)\, =\, 5\log(2)\, +\, 8$

. . . . .$x\left(3\, +\, \log(2)\right)\, =\, 5\log(2)\, +\, 8$

. . . . .$x\, =\, \frac{5\log(2)\, +\, 8}{3\, +\, \log(2)}$

Aqua Dragon
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Joined: Sat Apr 11, 2009 3:44 pm
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### Re: Solving for x in 10^(3x-8) = 2^(5-x)

huh.

...Wonder why I never thought of that. Guess I had expected a bit more... simplified version of x I guess. Thanks