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
That's exactly what you do.
I usually use natural logarithms...
$\displaystyle 3^{x + 1} = 7^{x - 4}$
$\displaystyle \ln{\left(3^{x + 1}\right)} = \ln{\left(7^{x - 4}\right)}$
$\displaystyle (x + 1)\ln{3} = (x - 4)\ln{7}$
$\displaystyle x\ln{3} + \ln{3} = x\ln{7} - 4\ln{7}$
$\displaystyle 4\ln{7} + \ln{3} = x\ln{7} - x\ln{3}$
$\displaystyle 4\ln{7} + \ln{3} = x(\ln{7} - \ln{3})$
$\displaystyle x = \frac{4\ln{7} + \ln{3}}{\ln{7} - \ln{3}}$.