Stumped....

[25thhour]

How Da fook do I solve these logs?

2^x-1 = (1/8)^3x

Ive tried logging both sides then expanding but It doesn't seem to work...


using LogBase5X = A and logbase5(1/5)x^2 find a formula with A in it/ Something like that.

[Reincarnate]

Re: Stumped....

http://www.wolframalpha.com

[dag12]

Re: Stumped....

taking the log of both sides should work. the bases don't really matter. (ie it can be log (base 10) or ln (base e))

if you decide to work with log,
2^(x-1) = (1/8)^(3x)
(x-1)*log2 = (3x)*log(1/8)
x*log2-log2 = (3x)*log(1/8)
x*log2-(3x)*log(1/8) = log2
x(log2 - 3*log(1/8)) = log 2
x = log 2 / (log2 - 3*log(1/8))
inputting this in your calculator should get you x = 0.1.

Alternatively, a little algebraic manipulation:
2^(x-1) = (1/8)^(3x)
2^(x-1) = (2^(-3))^(3x)
2^(x-1) = 2^(-9x)
since the exponents must be the same because you have the same base,
x-1 = -9x
10x = 1
x = 0.1

Also, if you use wolfram alpha, make sure you input your equation / expressions correctly.
2^x-1 means (2^x)-1, whereas I think you actually mean 2^(x-1). The parenthesis matter!

[25thhour]

Re: Stumped....

yep, I forgot the brackets. Sorry Thanks I finally figured it out on my own, took along time I was being a hurp durp with the basic math of dividing the logs.

[25thhour]

Re: Stumped....

I know, :P I was kinda in a rush.