How do you solve log_10(a^2-6)>log_10a?

1 Answer
Nov 10, 2016

graph{(y-ln(x^2-6)/ln10)(y-ln(x)/ln10)=0 [-1.785, 10.7, -1.773, 4.467]}

a>3

Explanation:

First ensure the logs exists:

{(a^2-6>0),(a>0):}=>a>sqrt6

Then cosider that log_10 is a growing function so the given relation is equivalent to

{(a>sqrt6),(a^2-6>a):}

{(a>sqrt6),(a^2-a-6>0):}

{(a>sqrt6),((a-3)(a+2)>0):}=>a>3