How to solve this problem: 2·3^x+1−4·3^x−2=150 (^ means power)?

　
If x+1 and x-2 are part of exponents, then you MUST use parentheses.

2·3^(x+1) − 4·3^(x−2) = 150
3^(x+1) − 2·3^(x−2) = 75
3^(x−2) (3^3 − 2) = 75
3^(x−2) · 25 = 75
3^(x−2) = 3
x − 2 = 1
x = 3

2(3^x) + 1 − 4(3^x) − 2 = 150
-2(3^x) = 150 - 1 + 2
-2(3^x) = 151
3^x = -151/2
3^x = -75.5
x = (ln(75.5) + i*pi*n) / ln(3), for any odd number n
x =~ 3.93599516486138420513487591 + 2.85960086738012726965214042*i*n, for any odd number n
(No real solutions.)
http://www.wolframalpha.com/input/?i=2%C2%B73%5Ex%2B1%E2%88%924%C2%B73%5Ex%E2%88%922%3D150

Or did you mean:
2(3^(x + 1)) - 4(3^(x - 2)) = 150
(3^(x + 1)) - 2(3^(x - 2)) = 75
3(3^x) - (2(3^x)/(3^2)) = 75
3(3^x) - (2/9)(3^x) = 75
(3 - (2/9))(3^x) = 75
(25/9)(3^x) = 75
3^x = 75 / (25/9)
3^x = 27
x = (ln(27) + 2*i*pi*n) / ln(3), for any integer n
x = 3 + ((2*i*pi*n)/ln(3)), for any integer n
x =~ 3 + 5.7192017347602545393042808406673*i*n, for any integer n
If x is a real number, then n = 0, so x = 3
http://www.wolframalpha.com/input/?i=2(3%5E(x+%2B+1))+-+4(3%5E(x+-+2))+%3D+150

let w = 3^x and solve : 2 (3) w - 4 (1/ 9 ) w = 150 for w....x = ln w / ln 3