# Homework Help: Exponential Equations

1. Sep 16, 2008

### JBD2

1. The problem statement, all variables and given/known data
Solve.
Q 1.) $$3^{3x}+3^{3x+2}=30$$

Q 2.) $$3^{2x}-12(3^{x})+27=0$$

2. Relevant equations

$$(a^{n}){b}=a^{nb}$$

$$a^{n}(a^{b})=a^{n+b}$$

3. The attempt at a solution

Q 1.) $$3^{3x}+3^{3x+2}=3^{1}+3^{3}$$

Q 2.) $$3^{2x}+3^{3}=12(3^{x})$$
$$3^{2x}+3^{3}=(3^{1}+3^{2})(3^{x})$$
$$3^{2x}+3^{3}=3{1+x}+3^{2+x}$$

2. Sep 16, 2008

### Defennder

Don't see how that helps. Solve for 3^(3x) first. Then use logarithms to find x.

Same technique as above. This time since it's a quadratic expression you have solve for 3^x using the quadratic formula. Then apply logs again.

3. Sep 16, 2008

### JBD2

Ignore on Question 2 my last proof of work, it's wrong, its supposed to be:

$$3^{2x}+3^{3}=3^{1+x}+3^{2+x}$$

And I would use logs but this question can supposedly be done without logs (the teacher hasn't shown us how to do them yet).

4. Sep 16, 2008

### snipez90

Hint

Q1. Let $$y = 3^{3x}$$. Can you write equation in terms of y?

5. Sep 16, 2008

### Melawrghk

Alrighty.

1. You can use the rule that states that ab * an = ab+n
33x + 33x+2=30
33x + 33x*32=30
Now you can isolate the 33x on the left hand side:
33x (1 + 9) = 30
33x=31
3x=1
x=1/3

Sorry, I'm not too sure about #2 yet. I'll keep trying it and I'll keep you posted if my mind comes up with something. :)

6. Sep 16, 2008

### JBD2

How did you go to this:
33x (1 + 9) = 30

I don't get how that works, can you explain it?

7. Sep 16, 2008

### JBD2

So you mean I'd have like $$3^{6x}(3^{2})=30$$?

EDIT: Nevermind I understand how you did it now, Same with my previous post I understand it, I'll work on the other question. Thanks.

By the way, Defennder, how did you know it was a quadratic expression?

Last edited: Sep 16, 2008
8. Sep 16, 2008

### Melawrghk

Yeah, of course.

33x+33x*32=30
You see how both terms on the left have 33x in them, right? So you can factor out that part. That will leave you 1 instead of the first term, and 32 instead of the second.
33x*(1+32)=30
32 is 9... Does that help at all?

9. Sep 16, 2008

### JBD2

Ya that makes sense, thanks. I just can't find any like terms in the second question now.

10. Sep 16, 2008

### JBD2

Ok I figured out the last question, and why it was in a quadratic, sorry for all the posts. Here's my solution:

$$3^{2x}-12(3^{x})+27=0$$
Like mentioned earlier, a quadratic:
$$x^{2}-12x+27$$
$$(x-9)(x-3)=0$$
$$(3^{x}-3^{2})(3^{x}-3^{1})$$
x=2, x=1