Please, show me how to do the steps for this equation, and the answer
2 to the power x which also has an exponent which is 2 = 32(2 which has an exponent 4x)
also, how do I put mathetical symbols into forums?
Hello, needhelpplease1!
Solve: .$\displaystyle 2^{x^2} \;=\;32^{4x}$
We have: .$\displaystyle 2^{x^2} \;=\;\left(2^5\right)^{4x}$
. . . . . . . .$\displaystyle 2^{x^2} \;=\;2^{20x} $
. - Then: . $\displaystyle x^2 \;=\;20x \quad\Rightarrow\quad x^2-20x\:=\:0 \quad\Rightarrow\quad x(x-20)\:=\:0$
Therefore: . $\displaystyle x\;=\;0,\;20$