2007-03-29 08:19:34 · 5 answers · asked by phenomena12 2 in Science & Mathematics ➔ Mathematics
log(9x)+3log(3x)=14
log(9)+log(x)+3(log(3)+log(x))=14
3^2=9, so...
2log(3)+log(x)+3log(3)+3log(x)=14
5log(3)+4log(x)=14
14-5log(3)=4log(x)
(14-5log(3))/4=log(x)
x=10^((14-5log(3))/4)
x=800.937
2007-03-29 08:30:07 · answer #1 · answered by Nick S 2 · 0⤊ 0⤋
Hi,
Log(9x) +3log(3x) =
log(9x)+log[(3x)^3]=
log(9x)+log27x^3=
log(9x*27x^3)=
log(243x^4)
If this equals 14 and the base for log is 10, then exponential form would be:
log(243x^4)=14
243x^4=10^14
x^4=(10^14)/243
Taking the 4th root of that big number, you get
x = 800.937138
To check it, substitute 800.937138 for x into
log(9x) +3log(3x) = log(9*800.937138) +3log(3*800.937138) AND, IT EQUALS 14!!!
I hope that helps!
2007-03-29 08:34:51 · answer #2 · answered by Pi R Squared 7 · 0⤊ 0⤋
Log 9X+ log (3x)^3 = 14
Log 9X+Log 27*X^3 = 14
Log 243 X^4 = 14
(10^14)/243= X^4
X= ( (10^14)/243)^1/4
2007-03-29 08:44:38 · answer #3 · answered by aziz wahba 1 · 0⤊ 0⤋
log(9x) + 3log(3x) = 14
log(9x) + log(27x^3) = 14
log(243x^4) = 14
243x^4 = 10^14
x = fourth root of [(10^14) / 243]
It's ugly but it works,
2007-03-29 08:27:15 · answer #4 · answered by tryzub91 3 · 0⤊ 0⤋
log (9x) + 3 log (3x) = 14
log (9x) + log (3x)^3 = 14
log (9x) + log (27x^3) = 14
log [ 9x • 27x^3] = 14
log (243x^4) = 14
243x^4 = 10^14
x^4 = 4.115 • 10^11
x = ±800.937
2007-03-29 08:27:09 · answer #5 · answered by Philo 7 · 0⤊ 0⤋