Quick Q

Your first step is correct, making all the bases 2.

You have gone badly wrong after that, did you forget all the rules of indices?

You should obtain a linear equation not a quadratic.

Just a bit stuck on this q;

$\frac{8^x}{4^{x-3}}=\frac{16}{2^{5x+2}}$

Am I right in thinking you first make them all the same number (2) so the top left one becomes 2^3x etc. Then cross multiply to get;

$15x^2-6x=8x-24$
then re-arrange to
$15x^2-14x+24=0$

I've been hinted that it should be then factorised into brackets, must the closest I can think of is
$(x-12)(x-2)=0$
which obviously works for most of it but ends up with x^2 instead of 15x^2. Can I just stick a 15 in front? or what!

+Rep to help

When you cross multiply and all the base of powers is 2 then
you should add the exponents not multiply.

You multiplied again!

$-6 + 4 \neq 24$

In LaTeX you can get a $\times$ by writing \times

$-6 + 4 = -2$

so

$2^{8x+2}=2^{2x-2}$

$6x+2=-2$

$6x = -4$

$x = -\frac{4}{6} = -\frac{2}{3}$
?