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.

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- July 28th 2011, 05:34 PMJamesBond16Logs 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.

- July 28th 2011, 05:34 PMProve ItRe: Logs and inequalities
Start by converting them to the same base...

- July 28th 2011, 05:40 PMJamesBond16Re: 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)

http://www2.wolframalpha.com/Calcula...=23&w=162&h=40 - July 28th 2011, 09:10 PMSammySRe: Logs and inequalities
Yes, is the correct solution for

I sure that**Prove It**assumed that what you meant by log_2x was , etc. The underscore symbol, "_" , generally means that what follows is a subscript. - July 28th 2011, 09:31 PMJamesBond16Re: Logs and inequalities
how how do i get x>67.7721 from this?

- July 28th 2011, 10:34 PMSammySRe: Logs and inequalities
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:

Changing the bases gives:

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

Use each side as the exponent with base e.

Solve it graphically.