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#1

$Old$ October 27th, 2007, 07:22 AM

ty2391 $ty2391 is offline$

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$Default$ Exponential/Log Problem

I think I figured out how to do this, but what would be the most efficient way to solve for x in this problem?

2^(4x-1) = 3^(1-x)

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#2

$Old$ October 27th, 2007, 08:29 AM

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Originally Posted by ty2391 $View Post$

2^(4x-1) = 3^(1-x)

You can't let the bases be equal, so apply logs:

$(4x-1)\log2=(1-x)\log3$

Now solve.

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#3

$Old$ October 27th, 2007, 10:16 AM

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Gotcha. So,

x= (log3 + log2) / (4 log2 + log3)
= 0.463

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