Intermediate Challenging Algebra

[Dragonballful23]

When is the following true?

3^x+1=9^x

[MAS1]

Originally Posted by Dragonballful23

When is the following true?

3^x+1=9^x

Did you mean 3^(x+1) = 9^x?

If so, then:

3^(x+1) = (3^2)^x
3^(x+1) = 3^(2x)
x+1 = 2x
x = 1

But the way the problem is originally written is much harder to solve.

(3^x) + 1 = 9^x
(3^x) + 1 = 3^(2x)
1 = 3^(2x) - 3^x
3^(2x) - 3^x - 1 = 0

Let a = 3^x

a^2 - a - 1 = 0
Then using the quadratic formula a = (1 + sqrt(5))/2 which is the golden ratio!

So 3^x = (1 + sqrt(5))/2
Taking log base 3 (log3) of both sides gives:
x = log3[(1 + sqrt(5))/2]

or using natural logs x = [ln(1 + sqrt(5)) - ln(2)]/(ln(3))

[Dragonballful23]

I apologize for leaving out the paretheses i made you do harder work But your answer is impeccable.

[eliseo]

It's 1 with the parentheses and 0.43801787946 without them.