Homework Help: Exponential Equations: Urgent Help

bluegirlbalance · Mar 4, 2007

I totally need urgent help on these exponential equations. I am so confused write now. I know that the bases for each equation has to be the same and that is x is squared you have to factor or use the quadratic formula to find the solution. I also know that the general steps to solve these equations are to make the bases on each side of the equation the same. You have to set the exponents equal to each other. Lastly, you solve for x. I am still confused, I could do the examples in the lesson, but no these practice problems. Help!!!!!!!!

1. 9^(2x)=27^(x-1)
2. 5^(n-1)=1/25
3. 25^x=5^(x^2-15)
4. (2^x)(4^(x+5))=4^(2x-1)
5. (sqrt 3)^(2x+4)=9^(x-2)

For the first equation, I think that x=-1. For the second equation, I think that n=1/2. For the other equations, I have no idea what to do.

arildno · Mar 4, 2007

Set your parentheses right about exponents before expecting help

cristo · Mar 4, 2007

Firstly, it would help if you put brackets around the exponents. For example, is question 1. 9^2x=27^x-1 or 9^2x=27^x-1? I presume it is the former, but I disagree with your answer. Show your work; what base would you choose to take on each side? What equation does this yield?

bluegirlbalance · Mar 4, 2007

cristo said: ↑

Firstly, it would help if you put brackets around the exponents. For example, is question 1. 9^2x=27^x-1 or 9^2x=27^x-1? I presume it is the former, but I disagree with your answer. Show your work; what base would you choose to take on each side? What equation does this yield?

okay, since the base has to be the same of both sides, I say that 3 would be the base for this equation. I solved the problem and I got -1/5. Is this right?

arildno · Mar 4, 2007

bluegirlbalance said: ↑

okay, since the base has to be the same of both sides, I say that 3 would be the base for this equation. I solved the problem and I got -1/5. Is this right?

Make your equation EXPLICIT at this point!
Otherwise, we can't help you along..

bluegirlbalance · Mar 4, 2007

arildno said: ↑

Make your equation EXPLICIT at this point!
Otherwise, we can't help you along..

1. 9^(2x)=27^(x-1)
Since the base has to be the same on both sides, I think that the base has to be changed to 3, making the equation read 3^(4)(2x)=3^(3)(x-1). You would set the exponents equal to each other: 8x=3x-1, which would give you x=-1/5. Is this the right answer for this problem, though?

arildno · Mar 4, 2007

No, it is not right, not your notation, and not your expression of 9, nor your contraction into an exponent equation.
We have:
[tex]9^{2x}=27^{(x-1)}, 9=3^{2},27=3^{3}\to(3^{2})^{2x}=(3^{3})^{x-1}\to3^{2*(2x)}=3^{3*(x-1)}[/tex]
yielding the exponent equation:
[tex]2*(2x)=3*(x-1)[/tex]

D H · Mar 4, 2007

No. It is easy to verify that [itex]9^{(2x)} \ne 3^{(4(2x)}[/itex]

bluegirlbalance · Mar 4, 2007

arildno said: ↑

No, it is not right, not your notation, and not your expression of 9, nor your contraction into an exponent equation.
We have:
[tex]9^{2x}=27^{(x-1)}, 9=3^{2},27=3^{3}\to(3^{2})^{2x}=(3^{3})^{x-1}\to3^{2*(2x)}=3^{3*(x-1)}[/tex]
yielding the exponent equation:
[tex]2*(2x)=3*(x-1)[/tex]

So I was right before, x= -1?

ZioX · Mar 4, 2007

bluegirlbalance said: ↑

So I was right before, x= -1?

Only if you believe -4=-6.

Do you understand what's going on here? We're saying that 9 can be represented by 3^2. Therefore, 9^x=(3^2)^x=3^(2x). Now the trick to solving these is to get them to the same base to get an 'exponent' equation.

bluegirlbalance · Mar 4, 2007

ZioX said: ↑

Only if you believe -4=-6.

Do you understand what's going on here? We're saying that 9 can be represented by 3^2. Therefore, 9^x=(3^2)^x=3^(2x). Now the trick to solving these is to get them to the same base to get an 'exponent' equation.

maybe it would help if I showed what I did to get x=-1 and then you could tell me what I am doing wrong.
9^(2x)=27^(x-1)
=3^(2)(2x)=3^(3)(x-1)
=4x=3x-1
=x=-1
So where did you get -4=-6 from?

D H · Mar 4, 2007

You are guessing (you original post) and being sloppy (your last post). What is [itex]3*(x-1)[/tex]?

cristo · Mar 4, 2007

bluegirlbalance said: ↑

maybe it would help if I showed what I did to get x=-1 and then you could tell me what I am doing wrong.
9^(2x)=27^(x-1)
=3^(2)(2x)=3^(3)(x-1)
=4x=3x-1
=x=-1
So where did you get -4=-6 from?

You expand 3(x-1) as 3x-1. This is not correct. It should be 3(x-1)=3x-3

arildno · Mar 4, 2007

Do you understand how to multiply out a parenthesis?

ZioX · Mar 4, 2007

bluegirlbalance said: ↑

maybe it would help if I showed what I did to get x=-1 and then you could tell me what I am doing wrong.
9^(2x)=27^(x-1)
=3^(2)(2x)=3^(3)(x-1)
=4x=3x-1
=x=-1
So where did you get -4=-6 from?

Okay, you're doing quite a few things wrong. Please don't be so liberal with your equality symbols. Break it up. Go the long way. Don't do shortcuts, this is where you're getting messed up.

9^(2x)=27^(x-1) iff (3^2)^(2x)=(3^3)^(x-1) iff 3^(4x)=3^(3x-3) iff 4x=3x-3 iff x=-3.

The key thing to realize here is that (a^n)^m=a^(nm). (Where these formulas make sense...of course :| )

bluegirlbalance · Mar 4, 2007

ZioX said: ↑

Okay, you're doing quite a few things wrong. Please don't be so liberal with your equality symbols. Break it up. Go the long way. Don't do shortcuts, this is where you're getting messed up.

9^(2x)=27^(x-1) iff (3^2)^(2x)=(3^3)^(x-1) iff 3^(4x)=3^(3x-3) iff 4x=3x-3 iff x=-3.

The key thing to realize here is that (a^n)^m=a^(nm). (Where these formulas make sense...of course :| )

Thank you Zio X. you have been a real help and I get it know.

ZioX · Mar 4, 2007

bluegirlbalance said: ↑

Thank you Zio X. you have been a real help and I get it know.

Don't thank me, thank arildno. He already told you precisely what I did. Except I gave you a final answer...which I'm kind of regretting.