Solve the given exponential equation without using logs?

Question

9^x = 3^x+3

remember without using logs please

Answer 1

9^x - 3^x = 3

I will try to come up with crude estimates of x.

If I ignored the 3 on the right side of the equation we have

3^2x/3^x = 1 which leads to 3^x =1 -----> first crude estimate is x = 0

Plugging that into the original equation we have 1 = 4, so lets try to get closer to the truth, substitute 3^x for 3 to have:

3^2x = 2*3^x ---------------> 3^x = 2, we would have to have an
0 Try x = 1/3 we get 3^1/3 = 1.442 ?= 2, that is closer than 1 = 4, so lets move up to x = 1/2
3 =? sqrt(3) + 3 = 4.73, so x is in (1/2,1) inverval, try x = 2/3

4.33 =? 5.08 4.33/5.08 = 0.85, compared to 3/4.73 = .63 , so x is in (2/3, 1) interval

So, I program my calculator to the equation 9^3x -3^x -3 = 0 and start plugging in values, I conclude x is about 0.759

x -------- f(x)

,759 is - 0.002
.76 -0.007
.77 0.99

Answer 2

Astrokid has the right plan, and I also do not see how you can progress without taking a log.

Is it possible you have the problem written down wrong? Maybe it's 9^x = 3^(x+3)?

Answer 3

3^2x=3^x + 3
3^x = y
y^2 = y+3
y^2 - y - 3 =0
y=(1+-sqrt(13))/2
3^x is positive, so y = (1+sqrt(13))/2
x=log(3) [{1+sqrt(13)}/2]