Question

I have to solve the equation "log(3x^2-3x-35)=0". I know that the answer is "x=-3 v x=4", but I'd like to know how to figure that out.

Answer 1

well raise each side of the equation as a power of the base of the log...

use the rule

\[e^{\ln(x)} = x\]

so I'll assume base e logs

\[e^{\ln(3x^2 -3x -35)} = e^0\]

which becomes

\[3x^3 -3x -35 = 1\]

hope you can solve the quadratic from here