wouldn't x have to be some sort of decimal?
2007-10-22 20:31:44 · 8 answers · asked by Conservative Jesus 1 in Science & Mathematics ➔ Mathematics
log of 15 to base 5 = 2x
log 15 / log 5 = 2x
x = log 15 / 2 log 5
x = 0.8413
[Check 5^(2 * 0.8413) = 15.0]
2007-10-22 20:41:12 · answer #1 · answered by Anonymous · 0⤊ 1⤋
5^2x = 15
Log.10(5^2x) = Log.10(15)
2x * Log.10(5) = Log.10(15)
2x * Log.10(10/2) = Log.10(5*3)
2x * Log.10(10) - Log.10(2) = Log.10(5) + Log.10(3)
2x * 1 - Log.10(2) = Log.10(5) + Log.10(3)
2x - Log.10(2) = Log.10(5) + Log.10(3)
2x = Log.10(5) + Log.10(3) + Log.10(2)
2x = Log.10(5*3*2)
2x = Log.10(30)
x = Log.10(30) / 2
Good luck.
2007-10-23 03:48:58 · answer #2 · answered by ¼ + ½ = ¾ 3 · 0⤊ 1⤋
2x log 5 = log 15
2x = log 15 / log 5
x = (1/2) [ log 15 / log 5 ]
2007-10-23 12:10:33 · answer #3 · answered by Como 7 · 0⤊ 0⤋
5^2x-15
25x=15
then u divide 25 fom both sides and then x =1.7
2007-10-23 04:36:04 · answer #4 · answered by montanajohnson429 1 · 0⤊ 0⤋
take logto any base e both sides
log(5^2x)=log(15)
2x=log15/log25 {as 5^2x=25^x }
x=1/2{(1.17)/(1.39)}
=1/2*.8418
=0.4209
2007-10-23 07:39:11 · answer #5 · answered by Einstein 1 · 0⤊ 0⤋
25^x=15. This is a logrithmic equation it has been too long since I did this type equation.
2007-10-23 03:41:56 · answer #6 · answered by Wylie Coyote 6 · 0⤊ 0⤋
Take logarithm to the base 5 on both sides :
log(5,5^(2x))=log(5,15)
2x=log(5,15)
2x=1.7
x=(1.7)/2
so x = 0.85
2007-10-23 03:42:51 · answer #7 · answered by Nobu 2 · 0⤊ 0⤋
take logarithm to the base 10 on both sides and then solve
2007-10-23 03:38:05 · answer #8 · answered by Anonymous · 1⤊ 1⤋