Logarithmic Question: Completed, Incorrect Answer

Hi,

How do I solve $\displaystyle 4(7^{x+2})=9^{2x-3} $?

If I take the natural logarithm of both sides:

$\displaystyle ln (4(7^{x+2})=ln(9^{2x-3}) $, and bringing the exponents in front of the expressions:

$\displaystyle (x+2)ln(4(7)) = (2x-3)ln(9) \to \frac{x+2}{2x-3}=\frac{ln(9)}{ln(28)} $

Where do I go from here or have I done this correctly?

$\displaystyle \frac{x+2}{2x-3}=0.6594 \to (2x - 3)(0.6594) = x + 2 \to 1.3188x -1.9782 = x + 2 \to .3188x = 3.9782 \to x = 12.4787 $

Where have I gone wrong?

Thanks!

Re: Logarithmic Question: Completed, Incorrect Answer

Quote:

Originally Posted by

**SC313** How do I solve $\displaystyle 4(7^{x+2})=9^{2x-3} $?

Oh come on?

$\displaystyle \ln(4\cdot 7^{x+2})=\ln(4)+(x+2)\ln(7)$

If you know anything about logarithms, you should know that.