how do I do that when there's no log to five :-/
2007-12-07 17:36:25 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let log be log to base two:-
8 - 6x = 2^5
6x = 8 - 2^5
6x = - 24
x = - 4
2007-12-07 23:14:58 · answer #1 · answered by Como 7 · 2⤊ 0⤋
log<2>(8-6x)=5
2^(log<2>(8-6x))=2^5
so tat the log will cancel out
8-6x=32 2^5=32
-6x=24
x=-4
log<2>(8-6*-4)=5
log<2>(8+24)=5
log<2>(32)=5
2^5=32 check
2007-12-08 01:51:04 · answer #2 · answered by tieu linh tinh 1 · 0⤊ 0⤋
log₂(8-6x)=5  â  8-6x = 2⁵
8-6x = 2⁵
-6x = 2⁵ -8
-6x = 32-8
-6x = 24
x = -4
2007-12-08 01:46:18 · answer #3 · answered by DWRead 7 · 0⤊ 0⤋
Your problem is the notation and the language.
Log, base 2, of (8-6x) = 5 is the logarithm form of this equation:
2^5 = (8-6x)
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The more general idea is
log form: log, base B, of (arg) = x is the same as
exponent form: B^x = arg
2007-12-08 01:50:31 · answer #4 · answered by answerING 6 · 0⤊ 0⤋
log<2>(8-6x)=5
2^5 = (8-6x)
32=8-6x
24=-6x
x=-4 .
2007-12-08 01:50:58 · answer #5 · answered by Murtaza 6 · 0⤊ 0⤋
i'm guessing you mean
log(base 2)(8 - 6x) = 5
2^5 = 8 - 6x
32 = 8 - 6x
-6x = 24
x = -4
2007-12-08 02:51:31 · answer #6 · answered by Sherman81 6 · 0⤊ 1⤋
2^5=8-6x
32=8-6x
6x=-24
x=-4
2007-12-08 01:41:22 · answer #7 · answered by someone else 7 · 0⤊ 0⤋