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?

Solve: . $2^{x^2} \;=\;32^{4x}$

We have: . $2^{x^2} \;=\;\left(2^5\right)^{4x}$

. . . . . . . . $2^{x^2} \;=\;2^{20x}$

. - Then: . $x^2 \;=\;20x \quad\Rightarrow\quad x^2-20x\:=\:0 \quad\Rightarrow\quad x(x-20)\:=\:0$

Therefore: . $x\;=\;0,\;20$

3. ## that's not the equation I meant :(

I'm sorry, I know that it was hard to understand, but that's not the equation I meant. This is what I need help with:

$2^ {x^2} \;=\;32(2^{4x})$

4. ## sorry, but now the equation I meant!

Originally Posted by Soroban

We have: . $2^{x^2} \;=\;\left(2^5\right)^{4x}$

. . . . . . . . $2^{x^2} \;=\;2^{20x}$

. - Then: . $x^2 \;=\;20x \quad\Rightarrow\quad x^2-20x\:=\:0 \quad\Rightarrow\quad x(x-20)\:=\:0$

Therefore: . $x\;=\;0,\;20$

SO SO sorry, I know it was confusing to read but what I meant was:

$2^ {x^2} \;=\;32(2^{4x})$