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 $
$\displaystyle (2x-3)lg2=lg32$
$\displaystyle x=4$