# solve for x???

• September 12th 2010, 04:02 AM
lollilikeslife
solve 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 :)
• September 12th 2010, 04:06 AM
Prove It
That's exactly what you do.

I usually use natural logarithms...

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

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

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

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

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

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

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