drawing a blank on this problem, if you can explain how to do it that would be much appreciated.
thanks
2007-12-11 05:54:18 · 6 answers · asked by chirch 1 in Science & Mathematics ➔ Mathematics
Use the fact that log(x/y) = log(x) - log(y). The rest should fall in to place.
2007-12-11 05:57:51 · answer #1 · answered by laurahal42 6 · 0⤊ 0⤋
log x - log(8-5x) = 2
Since we are subtracting logs:
log (x /(8-5x)) = 2
Raise to the power of 10 (assuming that this is log base 10 and not log base e)
x/(8-5x) = 10^2 = 100
Cross multiply
x = 800 - 500x
501x = 800
x = 800/501 As required.
Hope this helps
2007-12-11 05:58:55 · answer #2 · answered by highschoolmathpreparation 3 · 0⤊ 0⤋
don't let the logs fool you, okay
raise both sides (all terms) to power of 10
nuts, remember that log x - log y = log (x/y)
x / (8-5x) = 10^2 = 100
x = 100(8 - 5x)
501x = 800
x = 800/501 = 1.597
2007-12-11 06:05:51 · answer #3 · answered by Jim L 3 · 0⤊ 0⤋
Step 1: 5x-1 (+1) >9 (+1) Step 2: 5x (/5) >10 (/5) Step 3: x>2 Answer: x is greater than 2 or x>2
2016-05-23 02:03:20 · answer #4 · answered by ? 3 · 0⤊ 0⤋
log x - log(8-5x) = 2
=> log [x /(8-5x)] = 2
=> x /(8-5x) = 10^2
=> x /(8-5x) = 100
=> x = 100(8-5x)
etc
2007-12-11 05:59:00 · answer #5 · answered by harry m 6 · 0⤊ 0⤋
log 100 = 2, log m-log n= log(m/n)
log(x/(8-5x)) = log 100
x/(8-5x)=100
x =100(8-5x)
501x = 800
x = 800/501
2007-12-11 06:00:31 · answer #6 · answered by norman 7 · 0⤊ 0⤋