Solving equations with indeces (version 2)

Hello everyone,

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

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

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.

Re: Solving equations with indeces (version 2)

Quote:

Originally Posted by

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

$\displaystyle 9^{1-x}=3^{2-2x},~~27^{4+3x}=3^{12+9x},~~727=3^6$

Re: Solving equations with indeces (version 2)

These are all powers of three: $\displaystyle 27 = 3^3 ,\ 9 = 3^2\ ,\ 729 = 3^6$

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

Use the laws of indices to simplify