SOLUTION: I need your help with this problem please: Solve for X: 3^2x+3^x=6

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I need your help with this problem please: Solve for X: 3^2x+3^x=6       Log On


   



Question 930676: I need your help with this problem please:
Solve for X: 3^2x+3^x=6

Found 2 solutions by Alan3354, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!

Solve for X: 3^2x+3^x=6
---------
Sub for 3^x
3u^2 + 3u - 6 = 0
Solve for u
Sub back

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for X: 3^2x+3^x=6
3%5E%282x%29+%2B+3%5Ex+=+6
The easiest and least complicated method is to assign a VARIABLE to 3%5Ex. I will assign a, or replace 3%5Ex with a
Thus, 3%5E%282x%29+%2B+3%5Ex+=+6 becomes:
a%5E2+%2B+a+=+6 ---------- 3%5E%282x%29 is the same as %283%5Ex%29%5E2, leading to a%5E2
a%5E2+%2B+a+-+6+=+0 ------- Subtracting 6 from each side
(a - 2)(a + 3) = 0 ------- Factoring quadratic equation to solve for a
a - 2 = 0 OR a + 3 = 0
a = 2 OR a = - 3
a = 2
Since "a" was assigned to 3%5Ex, or was used to replace 3%5Ex, we can then say that a+=+2+=+3%5Ex
3%5Ex+=+2 ------------ Exponential form
log+%283%2C+2%29+=+x -------- Logarithmic form
x+=+log+2%2Flog+3, or highlight_green%28x+=+.63093%29
a = - 3
Since "a" was assigned to 3%5Ex, or was used to replace 3%5Ex, we can then say that a+=+-+3+=+3%5Ex
3%5Ex+=+-+3 ---------- Exponential form
log+%283%2C+-+3%29+=+x ------ Logarithmic form
IGNORE the above LOGARITHMIC FORM since a log CANNOT YIELD a value < 0.
You can do the check!!
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