Thread: Please help, solving exponential equations

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

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:

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

Originally Posted by Soroban

Hello, needhelpplease1!


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

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

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