2007-07-15 08:33:00 · 6 answers · asked by Bobby 1 in Science & Mathematics ➔ Mathematics
log2 (x+5)/(x-2)=3
2^3= (x+5)/(x-2)
8=(x+5)/(x-2)
8(x-2)=x+5
8x-16=x+5
7x=21
x=3
2007-07-15 08:37:49 · answer #1 · answered by haloobaloo 3 · 0⤊ 0⤋
log2 (x+5) - log2 (x-2) = 3 if the 2 is the base of the log then:
log2 (x+5) /(x-2) = 3 subtraction of logs is really division
(x+5) /(x-2) = 2^3 the base 2 is raised to power of 3
(x+5) = 8x-16 multiply both sides by (x-2)
21=7x combined terms and simplified
x=3 Answer
2007-07-15 08:40:15 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
log2(x+5-log2(x-2)=3
.
log2[ (x+5)/x-2)] =3
x+5 / x-2 = 2^3=8
.
x+5=8x -16
.5+16=8x-x
. 21= 7 x .....x=21/7=3
2007-07-15 08:45:24 · answer #3 · answered by Anonymous · 0⤊ 0⤋
log2(x+5)-log2(x-2)=3
log2(x+5)/(x-2)=3
(x+5)/(x-2)=2^3=8
x+5=8x--16
7x=21
x=3 ANS
2007-07-15 08:45:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Assume log means log to base 2 in the following:-
log [ (x + 5) / (x - 2) ] = 3
(x + 5) / (x - 2) = 2 ³
(x + 5) / (x - 2) = 8
x² + 3x - 10 = 8
x² + 3x - 18 = 0
(x + 6) (x - 3) = 0
x = 3 is acceptable answer.
2007-07-15 09:17:34 · answer #5 · answered by Como 7 · 0⤊ 1⤋
previous solutions have shown the perspective yet none has comprehensive off wisely. You do get from the quadratic equation x = a million.7085 or 12.2915 (4 dec. pl.) yet solutions ought to consistently be examined back interior the unique equation pretty the place logs are in contact. x = a million.7085 ----> log2 (x - 2) = log2 (-0.2915) that's undefined. x = 12.2915 ----> log2 (5 - x) = log2 (-7.2915) that's undefined. additionally log2 (a million - x) hence there are no longer any valid suggestions to this equation.
2016-12-10 13:03:29 · answer #6 · answered by ? 4 · 0⤊ 0⤋