i'm not sure how to do this
2007-09-23 03:31:49 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
a. 20/3
b. 6
c. 23/3
d. 7
2007-09-23 03:47:19 · update #1
Assume log to be to base 10 :-
log (2x) - log (x - 5) = 2
log [ 2x / (x - 5) ] = 2
2x / (x - 5) = 10 ²
2x = 100 (x - 5)
2x = 100 x - 500
98 x = 500
x = 500 / 98
x = 250 / 49
However, I strongly suspect you have presented the question incorrectly and you mean that all logs are to base 2. Assume this to be the case:-
log x - log (x - 5) = 2
log [ x / (x - 5) ] = 2
x / (x - 5) = 2 ²
x = 4x - 20
3x = 20
x = 20 / 3
OPTION a)
2007-09-23 06:25:41 · answer #1 · answered by Como 7 · 0⤊ 0⤋
log2x - log(x - 5) = log 10^2
2x/(x - 5) = 100
2x = 100x - 500
98x = 500
x = 500/98 = 5.1.
2007-09-23 10:45:16 · answer #2 · answered by man_mus_wack1 4 · 0⤊ 0⤋
OK
so first of all
2x>0---x>0
x>5
so x>5
2x/(x-5)=100
so x has to be different from 5
but that's already implemented in x>5
2x=100x-500
98x=500
so x=500/98 since x=5.10..... it can be taken as solution
2007-09-23 10:44:40 · answer #3 · answered by Alessandro 3 · 0⤊ 0⤋
log2x-log(x-5) = log10^2
2x/(x-5) = 100
2x = 10x-500
Solve for x,
x = 62.5
2007-09-23 10:36:34 · answer #4 · answered by sahsjing 7 · 0⤊ 1⤋