Expert: Sherman D. - 11/13/2006

Question
Solve for x

3^2x-3^x+1 + 2 = 0  

Answer
3^(2x) - 3^(x + 1) + 2 = 0
(3^2)^x - (3^x * 3^1) + 2 = 0
(3^x)^2 - 3(3^x) + 2 = 0

did you notice how (3^x)^2 - 3(3^x) + 2 = 0, looks a lot like

x^2 - 3x + 2 = 0

i just replaced 3^x with x, so

x^2 - 3x + 2 = 0
(x - 2)(x - 1) = 0
x = 1 or 2

but were not done yet. Since i replaced 3^x with x, you have to put 3^x back into the equation, like this

3^x = 1 or 3^x = 2
xln3 = ln1 or xln3 = 2
x = (ln1)/(ln(3)) or x = (ln(2))/(ln(3))
x = 0 or (ln2)/(ln3)

ANS : 0 or (ln2 / ln3)

if you go to www.quickmath.com, it will give you the same answer.

It will give you log instead of ln, but its the same value.

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Lets do this a way you can understand it

3^x = 1
log(3)1 = x
x = (log(1))/(log(3))
x = 0/(log(3))
x = 0

3^x = 2
log(3)2 = x
x = (log(2))/(log(3))

y^x = z
log(y)z = x

Using Logarithmic rule.