3^(x+1)=7^(x-4)

() are exponents :)

I dont really know what to do? I think we convert each side into a log: (x+1)Log3 = (x-4)Log7

But, I'm not sure? Please help! And provide work, I would really love to learn how to do this :)

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- Sep 12th 2010, 04:02 AMlollilikeslifesolve for x???
3^(x+1)=7^(x-4)

() are exponents :)

I dont really know what to do? I think we convert each side into a log: (x+1)Log3 = (x-4)Log7

But, I'm not sure? Please help! And provide work, I would really love to learn how to do this :) - Sep 12th 2010, 04:06 AMProve It
That's exactly what you do.

I usually use natural logarithms...

$\displaystyle 3^{x + 1} = 7^{x - 4}$

$\displaystyle \ln{\left(3^{x + 1}\right)} = \ln{\left(7^{x - 4}\right)}$

$\displaystyle (x + 1)\ln{3} = (x - 4)\ln{7}$

$\displaystyle x\ln{3} + \ln{3} = x\ln{7} - 4\ln{7}$

$\displaystyle 4\ln{7} + \ln{3} = x\ln{7} - x\ln{3}$

$\displaystyle 4\ln{7} + \ln{3} = x(\ln{7} - \ln{3})$

$\displaystyle x = \frac{4\ln{7} + \ln{3}}{\ln{7} - \ln{3}}$.