Question 670395
How do you figure out x in the equation log5(3x^2-1)=log5*2x
I assume you meant log5(3x^2-1)=log5(2x)
3x^2-1=2x
3x^2-2x-1=0
(3x+1)(x-1)=0
x=-1/3(reject,2x>0)
x=1