Answer:
x=2, x=-5
Step-by-step explanation:
to work more comfortably, the first thing we need to do is work the equation linearly, for that we take advantage of the property of logarithms that tells me that [tex]log(a^{b})=b*log(a)[/tex]
in this way, the equation remains as:
[tex]log(12^{x^{2} +5x-4 } )=log(12^{5x+6} ) \\ (x^{2} +5x-4 )*log(12)=(5x+6) *log(12)\\x^{2} +5x-4 = 5x+6[/tex]
Now we clear the equation so that it is of the form [tex]ax^{2} +bx+c=0[/tex]
[tex]x^{2} +5x-2x=6+4\\ x^{2} +3x-10=0[/tex]
finally, we apply the equation to solve second degree equations
[tex]x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}[/tex]
[tex]x = \frac{-3 \pm \sqrt {3^2-4*1*(-10)}}{2*1}\\ x = \frac{-3 \pm \sqrt {9+40}}{2}\\ x = \frac{-3 \pm 7}{2}[/tex]
x=2 and x=-5
Done
