Question
2logY=log2 + 3logX - 1
2^Y = 4^X - 2
2^Y = 4^X - 2
Answers
oobleck
using the rules of logs,
log y^2 = log(2*x^3/10)
y^2 = 1/5 x^3
2^y = 2^(2x-4)
y = 2x-4
combining those, you get
(2x-4)^2 = 1/5 x^3
x^3 - 20(x-2)^2 = 0
no easy roots there.
Maybe you should try using some parentheses, so we know just what you meant.
log y^2 = log(2*x^3/10)
y^2 = 1/5 x^3
2^y = 2^(2x-4)
y = 2x-4
combining those, you get
(2x-4)^2 = 1/5 x^3
x^3 - 20(x-2)^2 = 0
no easy roots there.
Maybe you should try using some parentheses, so we know just what you meant.