Solve log2(x) that's base 2 and base 3.I tried using change of base w/ ln and then multiplying by ln2ln3 but I'm not sure where to...

First, you need to create matching bases to both
logarithms:

log a (x) = log b (x)/log b
(a)

log3(x) = log2(x)/log2 (3) => log2(x) =
log3(x)*log2 (3)

The inequality will
become:

log3(x)*log2 (3) -
log3(4x-5)

But log2 (3) = lg3/lg2 =
0.4771/0.3010=1.585

1.585log3(x) - log3(4x-5) >
1

log3 (x^1.585) - log3(4x-5) >
1

Since the bases are matching, we'll use quotient
rule:

log3 (x^1.585/(4x-5)) >
1

We'll write 1 as log3
(3)

x^1.585/(4x-5) >
3

(x^1.585 - 12x + 15)/(4x-5) >
0

Both numerator and denominator must be positive or
negative, to keep the fraction positive.

The denominator is
positive for x>5/4