2^(4x-3)=32?

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2007-06-28 18:41:09 · 5 answers · asked by eric m 1 in Science & Mathematics ➔ Mathematics

32 = 2^5

2^(4x-3) = 2^5
4x-3=5
4x=8
x=2

2007-06-28 18:45:35 · answer #1 · answered by gudspeling 7 · 0⤊ 0⤋

2^4x-3=32
<=> 4x-3=Log base 2 of 32
<=> 4x-3=5 (2^5=32)
<=> 4x=5+3
<=> 4x=8
<=> x=8/4
<=> x=2

2007-06-29 02:55:56 · answer #2 · answered by God_Of_War 2 · 0⤊ 0⤋

2^(4x-3)=32
Put log on both sides:
log 2^(4x-3)= log 32
(4x-3)*log 2=log 32
4x-3=log32/log2
4x-3=5
4x=8
x=2

2007-06-29 01:48:21 · answer #3 · answered by nivik 3 · 0⤊ 0⤋

2^(4x-3)=32
2^(4x-3)=2^5
therefore 4x-3 = 5
4x = 8
x = 8/4
x = 2 #

2007-06-29 03:03:27 · answer #4 · answered by wanda 3 · 0⤊ 0⤋

take log base 2:-
(4x - 3) log 2 = log 32
4x - 3 = 5
4x = 8
x = 2

2007-07-02 17:34:55 · answer #5 · answered by Como 7 · 0⤊ 0⤋