Thread: Solving equations with indeces

1. Solving equations with indeces

Hello everyone,

I am having trouble solving the very simple equation:

$\displaystyle 2^x(2^{x-3})=32$

I can see that $\displaystyle x=4$ is the right answer, but how could I go about showing some working for this?

2. Re: Solving equations with indeces

Originally Posted by mabentley
Hello everyone,

I am having trouble solving the very simple equation:

$\displaystyle 2^x(2^{x-3})=32$

I can see that $\displaystyle x=4$ is the right answer, but how could I go about showing some working for this?
$\displaystyle 2^x(2^{x-3})=32$
$\displaystyle 2^{x+x-3}=2^5$
by comparing indices,
2x-3=5
x=4

alternatively,
$\displaystyle 2^x(2^{x-3})=32$
take lg of both sides,
$\displaystyle lg 2^{x+x-3}=lg32$
$\displaystyle (2x-3)lg2=lg32$
$\displaystyle x=4$