# Logarithmic Question: Completed, Incorrect Answer

• January 9th 2013, 05:21 PM
SC313
Hi,

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

If I take the natural logarithm of both sides:

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

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

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

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

Where have I gone wrong?

Thanks!
• January 9th 2013, 06:00 PM
Plato
Re: Logarithmic Question: Completed, Incorrect Answer
Quote:

Originally Posted by SC313
How do I solve $4(7^{x+2})=9^{2x-3}$?

Oh come on?
$\ln(4\cdot 7^{x+2})=\ln(4)+(x+2)\ln(7)$

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