Question

Solve for x:
2^(x-x^2)=1/8^(x)

So far I'm at:
x-x^2ln2=-xln8

So do I just divide the (x-x^2) by x and ln8 by -ln2?

Answer 1

Im supposed to have two different x values as my answer, but am I even doing it right?

Answer 2

Hi there :)\[\Large\rm 2^{x-x^2}=\left(\frac{1}{8}\right)^{x}\]Don't use logs.
That's a sloppy approach.
We want to write our `right side` as a power of 2.
8 is 2 to the 3rd power,\[\Large\rm 2^{x-x^2}=\left(\frac{1}{2^3}\right)^{x}\]We'll bring the 2 up to the numerator using rule of exponents,\[\Large\rm 2^{x-x^2}=\left(2^{-3}\right)^{x}\]Another rule of exponents let's us multiply the -3 and x,\[\Large\rm 2^{x-x^2}=2^{-3x}\]

Answer 3

From here, since our bases are equivalent,
it means our exponents must be equivalent as well,\[\Large\rm x-x^2=-3x\]Solve for x! :)