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.
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)
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.
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.