How would I solve the following problem?
4^3x-7 =5
Thanks
2007-04-25 22:58:24 · 6 answers · asked by Sean M 2 in Science & Mathematics ➔ Mathematics
If u meant by 4^(3x-7)=5, then
log4^(3x-7)=log5
(3x-7)log4=log5
(3x-7)= log5/log4
3x-7=1.161
3x=8.161
x=2.72
If u meant by 4^3x -7=5, then
4^3x=5+7
log4^3x=log12
(3x)log4=log12
3x= log12 / log4
3x=1.792
x=0.597
2007-04-25 23:06:51 · answer #1 · answered by (^InLove^) 3 · 0⤊ 0⤋
If it's (4^3x)-7 = 5
4^3x=12
3x log4=log12
3x = log12/log4
x=log12/3log4
If it's 4^(3x-7) = 5
(3x-7)log4=log5
3x-7=log5/log4
3x=(log5/log4)+7
x=log5/3log4 + 7/3
2007-04-25 23:09:05 · answer #2 · answered by solver 3 · 0⤊ 0⤋
4^3x - 7 = 5
4^3x = 12
3x(log 4) = log 12
3x = (log 12)/(log 4)
3x = 1.7924813
x = 0.5974937
2007-04-25 23:06:52 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
4^(3x) = 12
3x.log 4 = log 12
x = (log 12 / log 4) / 3
x = 0.597
2007-04-26 00:23:16 · answer #4 · answered by Como 7 · 0⤊ 0⤋
4^3x = 12
3xlog(4) = log(12)
3x = log(12)/log(4) = log(3)
x = log(3)/3
2007-04-25 23:05:26 · answer #5 · answered by S 2 · 0⤊ 0⤋
4^(3x) - 7 = 5
Add 7:
4^(3x) = 12
Take logarithms:
3x log(4) = log(12)
Divide by 3log(4):
x = log(12) / 3log(4).
2007-04-25 23:03:24 · answer #6 · answered by Anonymous · 0⤊ 0⤋