[Highschool - Math] Having troubles Solving Exponential Equations

[og4lif]

Hello. I'm having difficulties solving this particular Exponential Equation and according to my understanding it involves logarithms.

3^2x - 6(3^x) + 9 = 0

I'm not interested in the answer. I already know that the answer is one but I am having difficulties manipulating this equation. I would really appreciate it if someone could show me what I should do because I have lots of homework with similar questions.

Below I have provided some work I attempted on this question.

3^2x - 6(3^x) + 9

log 3^2x - [log 6 + log 3^x] = - log 9

2x log 3 - log 3^x = log 6 - log 9

x[log 3^2 - log 3] = log 6/9

Eventually, If i continue I end up with the wrong answer.

[devonin]

I edited your thread title to follow the explicit thread titling requirements in the rules. That's your one warning to remember to follow directions.

[flipsta_lombax]

Hello. I'm having difficulties solving this particular Exponential Equation and according to my understanding it involves logarithms.

3^2x - 6(3^x) + 9 = 0

I'm not interested in the answer. I already know that the answer is one but I am having difficulties manipulating this equation. I would really appreciate it if someone could show me what I should do because I have lots of homework with similar questions.

Below I have provided some work I attempted on this question.

3^2x - 6(3^x) + 9

log 3^2x - [log 6 + log 3^x] = - log 9

2x log 3 - log 3^x = log 6 - log 9

x[log 3^2 - log 3] = log 6/9

Eventually, If i continue I end up with the wrong answer.

The way I see it:
You may have brought your x down wrong?

Instead of : log 3^2x - [log 6 + log 3^x] = - log 9

I see it as : 2x log 3 - 6x log 3 = - log 9
-4x log 3 = - log 9
x = - log 9 / - 4 log 3


I am not quite sure, because we are doing exponential equations as well, but they are quite different from this.

[Jtehanonymous]

Wait I'm sorry, nevermind. -.- Scratch everything I just said..

[og4lif]

Thanks a lot buddy I got it. Let me post my work in a second for you to look over

2x log 3 - log 3^6x = -log 9

2x log 3 - 6xlog 3 = - log 9

2x (log 3 - 3log 3) = -log 9

2x (log 3/9) = - log 9

2x = - log 9 / log 1/3

2x = 2

x = 1 =) Thanks for the help flipsta_lombax and Jtehanonymous.

[flipsta_lombax]

I don't agree with you jtehanonymous. I just don't see where you need to take the log of 6 when all you need to take the log of is that 3^x.

Oh btw, my answer:
x = - log 9 / - 4 log 3
x = log 9 / 4 log 3
x = .5


The logistics seem quite reasonable on my behalf.

[Jtehanonymous]

I don't agree with you jtehanonymous. I just don't see where you need to take the log of 6 when all you need to take the log of is that 3^x.

The logistics seem quite reasonable on my behalf.

I realized that when I looked over my post again, and that's why I felt stupid and edited it. XD

[og4lif]

.5 doesn't seem to work if i place it back into the equation.

But yeah thanks again guys. So does this thread get locked up now?

[Jtehanonymous]

2x log 3 - log 3^6x = -log 9

2x log 3 - 6xlog 3 = - log 9

2x (log 3 - 3log 3) = -log 9

2x (log 3/9) = - log 9


Wouldn't that part be 2x (Log 3/27) = -log9?

I dunno, I thought the 3log3 would make a log 27

Edit : 2x (Log 1/9) = -log9

2x = -log9/log1/9 (Which equals 1)

2x = 1

x = 1/2

So I agree with flip. XD

[NFD]

...What math level is this? Algebra II? I'm in Accelerated Geometry, I can't figure it out, x.x

[devonin]

Yay the successful conclusion of our inaugural thread.

[Master_of_the_Faster]

Hello. I'm having difficulties solving this particular Exponential Equation and according to my understanding it involves logarithms.

3^2x - 6(3^x) + 9 = 0

I'm a wizard at this. First thing is first.

If you look carefully, this question is somewhat different from your average exponential equation question in that it is similar to a quadratic equation.

Picture this: instead of 3^x, think of it as x and 3^2x as X^2. Everything in this equation should look like (X^2)-6(x)+9=0
If you solve the quadratic properly, you should factor and get (x-3)(x-3)=0

Now think of it in terms of the actual equation. You should get factors of
((3^x)-3)((3^x)-3)=0

From here, you should solve for (3^x)-3=0 in the usual fashion by the following:
1. adding 3 to both sides to make this: (3^x)=3
2. do log base 3 on both sides to look like this: Log3(3^x)=Log3(3)
3: Your x value should be equal to Log3(3) or 1

Always check your answer by plugging it into the original equation.

[sumzup]

Yay the successful conclusion of our inaugural thread.
Master of the Faster: Conclusion usually means no bumping 2 months after the fact.

Edit: I suppose, though, that adding to the discussion doesn't really do much harm, especially in such a slow moving subforum. Whatever.