Okay, this is the equation: 25^{x}=5^{x*x -3} * x. Solve for x.
Hello, waldol1!
I assume you have a typo . . .
$\displaystyle \text{Solve for }x\!:\;25^x \:=\:5^{x^2-3x}$
We have: .$\displaystyle (5^2)^x \:=\:5^{x^2-3x} \quad\Rightarrow\quad 5^{2x} \:=\:5^{x^2-3x}$
. . . . . . . . $\displaystyle 2x \:=\:x^2-3x \quad\Rightarrow\quad x^2 - 5x \:=\:0 \quad\Rightarrow\quad x(x-5) \:=\:0$
Therefore: .$\displaystyle x \:=\:0,\:5$