Working with logarithms?

I know the log rules and I am comfortable changing bases and working with log equations, but this question is confusing:

log3(23x)=log9(6x219x2)

Thanks!

1 Answer
Feb 26, 2018

There are no real solutions to this problem

Explanation:

Using logab=logalogb we can rewrite the equation
log3(23x)=log9(6x219x2)
as

log(23x)log3=log(6x219x2)log9
log(23x)×log9log3=log(6x219x2)
2log(23x)=log(6x219x2)
log(23x)2=log(6x219x2)
(23x)2=6x219x2
412x+9x2=6x219x2
3x2+7x+6=0

This quadratic equation has a discriminant of
724×3×6=4972=23<0
so there are no real roots.