Hi, I've been attempting this problem for a while now, and I can't see how they got the solution.
Problem: 3^(x-1)=2^(2x-1)
Solution: (log 3-log 2) / (log 3-2log 2)
Can anyone show me the method for solving such an equation?
Another slightly different way of doing it:
3^(x - 1) = 2^(2x - 1)
log [3^(x - 1)] = log [2^(2x - 1)]
(x - 1) log 3 = (2x - 1) log 2
x log 3 - log 3 = 2x log 2 - log 2
x log 3 = 2x log 2 - log 2 + log 3
x log 3 - 2x log 2 = log 3 - log 2
x(log 3 - 2 log 2) = log 3 - log 2
x = {log 3 - log 2)/(log 3 - 2 log 2)
Hello, ima9rd!
I'll solve it the traditional way . . .
Problem: . 3^(x-1) .= .2^(2x-1)
Solution: .[log(3) - log(2)] / [(log(3) -2·log(2)]
Take the log of both sides:
. . . . . . . . . . . log[3^(x-1)] .= .log[2^(2x-1)]
. . . . . . . . . . . (x-1)·log(3) .= .(2x-1)·log(2)
. . . . . . . . x·log(3) - log(3) .= .2x·log(2) - log(2)
. . . . . . x·log(3) - 2x·log(2) .= .log(3) - log(2)
Factor: .x[log(3) - 2·log(2)] .= .log(3) - log(2)
. . . . . . . . . . . . . . log(3) - log(2)
Therefore: .x . = . -------------------
. . . . . . . . . . . . . log(3) - 2·log(2)