hint... first rewrite each base (27 and 3) to the power of 3 then use the index rule.
please help
2007-04-18 21:53:15 · 5 answers · asked by Joanne P 1 in Science & Mathematics ➔ Mathematics
3^(12x-6)=3^(3-x)
12x-6=3-x
x=9/13
2007-04-18 21:57:38 · answer #1 · answered by iyiogrenci 6 · 0⤊ 0⤋
Take log base 3 of both sides:-
(4x - 2).log 27 = (3 - x).log 3
(4x - 2). 3 = (3 - x).1
12x - 6 = 3 - x
13x = 9
x = 9 / 13
2007-04-18 22:34:00 · answer #2 · answered by Como 7 · 0⤊ 0⤋
3 to the power of 12x-6 = 3 to the power of 3-x
so
3-x=12x-6
so x=9/13
check:3-9/13 =12*9/13-6
30/9=30/9
2007-04-18 22:09:52 · answer #3 · answered by john s 1 · 0⤊ 0⤋
3^3(4x-2)=3^3-x
now you can solve the exponents
3(4x-2)=3-x
solve for x and voila
2007-04-18 21:57:23 · answer #4 · answered by Anonymous · 0⤊ 0⤋
a)27=3^3=>27^(4x-2)=(3^3)^(4x-2)=3^[3*(4x-2)]=3^(12x-6)
b)3^(12x-6)=3^(3-x)
=>12x-6=3-x /+x+6
<=>13x=9
=>x=9/13
next time please give the formula
2007-04-18 22:00:10 · answer #5 · answered by Bladvak 3 · 0⤊ 0⤋