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

Question

1781 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-10T15:22:25

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

#a>3#

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#