6 times 4^2x-3 = 37
Thank you.
2006-10-10 06:31:37 · 5 answers · asked by pnoiz1 2 in Science & Mathematics ➔ Mathematics
I assume you mean 6(4^(2x-3)) = 37.
Divide by 6:
4^(2x-3) = 37/6
Take natural log of both sides:
ln 4^(2x-3) = ln (37/6)
Use property ln x^a = a ln x:
(2x-3) ln 4 = ln (37/6)
Divide:
2x-3 = ln(37/6) / ln 4
Solve for x:
x = (3 + ln(37/6) / ln 4)/2
2006-10-10 06:39:26 · answer #1 · answered by James L 5 · 0⤊ 0⤋
do you mean 4^(2x-3)=37?
If the problem is how you stated it, then the order of operations would be:
16x - 3 = 37
x = 40/16=5/2
2006-10-10 15:03:22 · answer #2 · answered by Melody 3 · 0⤊ 0⤋
take log
log 6+(2x-3)log4=log37
2x-3 log4=log37-log6
2x=3log4+log37-log6
x=[log(37*4^3)/6]/2
2006-10-10 13:51:22 · answer #3 · answered by raj 7 · 0⤊ 0⤋
exp (log(((37/6) + 3 ))/2)
2006-10-10 13:35:38 · answer #4 · answered by gjmb1960 7 · 0⤊ 0⤋
6*[4]^2x-3=37
[4]^2x-3=37/6
2x-3=lg[37/6]/lg4
2x-3=lg[37/6-4]=lg[(37-24)/6]
2x-3=lg[13/6]=lg2.16=.3344
2x=3.3344
x=1.6672
2006-10-10 13:44:58 · answer #5 · answered by openpsychy 6 · 0⤊ 0⤋