College Mathematics

[studyhard]

Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?

[DixieGirl]

Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?

Did you mean
8^x = 15
8^y = 25
8^(2x+y) = ??

I couldn't figure it out either, but I posted this on another forum and they said,

"since 8x = 15 ... 82x = (8x)2 = 152.

82x+y = 82x*8y = 152*25"

Hope this helps!:o

[wannabeit]

Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?

I think (x =15 and (y)=25, therefore 8 x (2 x 15+ 25)
(30 +25) = 55
8 x 55 = 440

[sgloer]

I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so:
8^(2x + y) is the same as
8^(2x) + 8^(y) which is the same as
8^x + 8^x + 8^y
= 15 + 15 + 25
= 55

If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly.

[tryingtesc]

Ok...There are two ways to solve this problem...One is the use of Logs to find an unknown exponent. For example:
8^x=15
= log 15 base 8 = x (re-write in log form)
=log 15 / log 8 (divide)
=1.3022968652 (this is the "x" exponent)

8^y=25
=log 25 base 8 = y
=log 25 / log 8
= 1.54795206326 (this is the "y" exponent)

Plug your now Known exponents into your equation:
8^2x+y = 8^2(1.3022968652)+(1.54795206326) = 5624.9999 = 5625

Second Way:
They made this easy because of the same base 8.
8^x=15 , 8^y=25, 8^2x+y=??
8^2x+y = 8^2x * 8^y (re-write by seperating exponents)
8^y is the easy part, we know that is 25
8^2x is not the same of course as 8^x (which we know is 15) so what makes it different? Yes the "2". Well you squared the left side of the equation so you have to square the right side: 8^2x = 15^2. Now you know 8^2x = 225 (15 squared).
So put the two together: 8^2x * 8^y is now the same as 225*25 = 5625

^ = raised to that exponent
* = multiplication

I hope this helps. It is a tough problem to explain on a message board.
Best regards,
Jason

[tryingtesc]

I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so:
8^(2x + y) is the same as
8^(2x) + 8^(y) which is the same as
8^x + 8^x + 8^y
= 15 + 15 + 25
= 55

If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly.

You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5.
You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed.

I hope this helps with your reverse engineering because I am not familiar with that method.

Best Regards,
Jason

[radyogyrl]

For this particular problem, you will square 15 which will give you 225 then mulitiply 225 by 25 which will give you 5625. I had a hard time figuring this one out when I was studying for the college math clep. I just played with it until I got the correct answer.

[sgloer]

You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5.
You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed.

I hope this helps with your reverse engineering because I am not familiar with that method.

Best Regards,
Jason

Yeah, that looks right. Thanks Jason--I couldn't remember exactly how the rule went.