2007-03-23 19:12:42 · 5 answers · asked by John S 1 in Science & Mathematics ➔ Mathematics
since 27 = 3^3 and 9 = 3², we can write the problem as
(3^3)^(6x-5) = 3²
3^(18x-15) = 3²
18x - 15 = 2
18x = 17
x = 17/18
2007-03-23 19:21:35 · answer #1 · answered by Philo 7 · 0⤊ 1⤋
let's take log (base 9) of both sides
(6x-5) log 27 = log 9.But 27 =9* sqrt9 so log27 = 1+1/2=3/2
So 3/2(6x-5)=1 6x-5 = 2/3 and x= 17/18
2007-03-24 03:39:07 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
Take log base 3 of both sides
(6x - 5).log 27 = log 9
6x - 5 = log(9) / log(27) = 2/3
6x = 17/3
x = 17/18
2007-03-23 23:13:35 · answer #3 · answered by Como 7 · 0⤊ 0⤋
we know 27 can be written as 3^3 and 9 as 3^2
then,
3^3(6x-5)=3^2
when bases are equal ,then powers must be equal
18x-15=2
18x=17
x=17/18
2007-03-23 22:56:38 · answer #4 · answered by satwik 2 · 0⤊ 0⤋
27 = 3^3 and 9 = 3^2
(3^3)^(6x-5) = 3^2
3^(18x - 15) = 3^2
18x - 15 = 2
12x = 17
x = 17/18
2007-03-23 19:21:15 · answer #5 · answered by 7 · 1⤊ 1⤋