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by **Aqua Dragon** on Tue May 05, 2009 5:30 am

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.

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by **stapel_eliz** on Tue May 05, 2009 11:07 am

Aqua Dragon wrote: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". :wink:

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

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

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

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

:wink:

by **Aqua Dragon** on Wed May 06, 2009 3:37 am

huh.

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

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