Exponential Equations: Solve for x with Step-by-Step Solutions

[JBD2]

Homework Statement

Solve.
Q 1.) 3^{3x}+3^{3x+2}=30

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

Homework Equations

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

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

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}

 

[Defennder]

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

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

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}

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.

 

[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).

 

[snipez90]

Hint

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

 

[Melawrghk]

Alrighty.

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

 

[JBD2]

Alrighty.

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

How did you go to this:
3^3x (1 + 9) = 30

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

 

[JBD2]

Hint

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

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?

 

[Melawrghk]

How did you go to this:
3^3x (1 + 9) = 30

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

Yeah, of course.

3^3x+3^3x*3²=30
You see how both terms on the left have 3^3x 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.
3^3x*(1+3²)=30
3² is 9... Does that help at all?

 

[JBD2]

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

 

[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