1. ## Logs and inequalities

My study guide states that log_2x-log_3(4x-5)>1 is x>67.7721 but I have yet to rap my head around logs. If someone could reveal to me the process behind this it would greatly appreciated.

2. ## Re: Logs and inequalities

Start by converting them to the same base...

3. ## Re: Logs and inequalities

Would be to much trouble to ask you to show me?

is it okay to leave it as a inequality?

Since the base is the same, you can actually make this log (2x/(3(4x-5))) >1 or
log (2x/(12x-15)) >1. And this simply means that (2x/12x-15)>10.

Temporarily assuming that 12x-15 is positive (x>1.25), you can multiply each side by this and get
2x>120x -150. Then 2x +150>120x, then 150>118 x. Then x<150/118 (but still greater than 1.25).

Temporarily assuming that 12x-15 is negative (x<1.25), you can do the same stuff with the sign pointing the other way, 150<118x , then x>150/118, but this is a contradiction.

So I'm sticking with 1.25<x<150/118 (this fraction is approx 1.271 btw, so a narrow value range)

4. ## Re: Logs and inequalities

Yes, $\displaystyle \frac{5}{4}<x<\frac{75}{59}\,,$ is the correct solution for $\displaystyle \log_{10}(2x)-\log_{10}(3(4x-5))>1$

I sure that Prove It assumed that what you meant by log_2x was $\displaystyle \log_2(x)$, etc. The underscore symbol, "_" , generally means that what follows is a subscript.

5. ## Re: Logs and inequalities

how how do i get x>67.7721 from this?

6. ## Re: Logs and inequalities

Originally Posted by JamesBond16
how how do i get x>67.7721 from this?
I tried to solve it assuming that the bases were 2 & 3 as Prove It reasonably assumed. This does give the approximate answer: x>67.7721.

Assume: $\displaystyle \log_{2}(x)-\log_{3}(4x-5)>1\,.$

Changing the bases gives: $\displaystyle \frac{\ln(x)}{\ln(2)}-\frac{\ln(4x-5)}{\ln(3)}>1\,.$

Multiply both sides by ln(2) and ln(3).

Use each side as the exponent with base e.

Solve it graphically.