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)

Postby Aqua Dragon » 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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stapel_eliz
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Postby stapel_eliz » 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:

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

Postby Aqua Dragon » 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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