# Solve log2(x) - log3(4x-5) > 1 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 go from there. ln(3)ln(x) - ln(2)ln(4x-5) >...

Solve log2(x) - log3(4x-5) > 1

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 go from there.

ln(3)ln(x) - ln(2)ln(4x-5) > ln(2)ln(3) and then? Or is there another way.

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After playing with this for a while I do not see anyway to solve for x.

If you are in a Calculus class there is a way to approximate this root using Newton's Method.

I graphed y=log_2(x) - log_3(4x-5) - 1 and zeroed in on where this crossed the x axis. I get x `~~` 1.395243. So log_2(x) - log_3(4x-5) > 1 when x < 1.3905243

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