# Solving equations with indeces (version 2)

• Sep 4th 2011, 09:35 AM
mabentley
Solving equations with indeces (version 2)
Hello everyone,

I have come undone on solving another equation with indices in:

$\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$

I've a feeling it involves simplifying the numerator and denominator of the second half so that I can multiply with the first half, but I'm doing something wrong.

Help much appreciated.
• Sep 4th 2011, 09:43 AM
Plato
Re: Solving equations with indeces (version 2)
Quote:

Originally Posted by mabentley
$\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$

$9^{1-x}=3^{2-2x},~~27^{4+3x}=3^{12+9x},~~727=3^6$
• Sep 4th 2011, 09:44 AM
e^(i*pi)
Re: Solving equations with indeces (version 2)
These are all powers of three: $27 = 3^3 ,\ 9 = 3^2\ ,\ 729 = 3^6$

Hence the question is: $\dfrac{3^{5x+2}}{3^{2(1-x)}} = \dfrac{3^{3(4+3x)}}{3^6}$

Use the laws of indices to simplify