Find Y when AxY=B, BxY=C, and B is unknown

[1plus1is10]

How do I:
Find Y when AxY=B, BxY=C, and B is unknown?
(A and C are known)

Example: If A=100 and C=169 then Y=1.3 and B=130

I assume I use log or pow but cannot figure it out.
Thanks.

 

[Mark44]

How do I:
Find Y when AxY=B, BxY=C, and B is unknown?
(A and C are known)

Example: If A=100 and C=169 then Y=1.3 and B=130

I assume I use log or pow but cannot figure it out.
Thanks.

No, logs are not needed.
Just solve for y in the first equation, then solve for y in the second equation, and equate the two expressions for y. If you know a and b, then you should be able to find c, and from that, you can find y.

Tips: Algebra equations typically use lower-case letters for variables. Also, don't use x for multiplication, especially in equations that already involve x (which your equations don't). Instead of writing AxY and BxY, just write AY and BY or ay and by.

 

[1plus1is10]

Okay... If B=A*Y and B=C/Y then C/Y=A*Y

Solution One:
A=(C/Y)/Y = C/(Y*Y) = C*1/(Y*Y) = 1/(Y*Y)*C
A/C=1/(Y*Y)
(A/C)/(Y*Y)=1
(A/C)*1/(Y*Y)=1
1/(Y*Y)=1/(A/C)
I'm clueless what happens next

Solution Two:
C=(A*Y)*Y = A*(Y*Y) = (Y*Y)*A
C/A=Y*Y
sqrt(C/A)=Y

Test:
sqrt(169/100)=1.3
Yeah! I did it!

Thanks Mark for pointing the way.

 

[1plus1is10]

PS... sorry about using Caps

 

[1plus1is10]

A/C=1/(Y*Y)
(A/C)*(Y*Y)=1
(Y*Y)*(A/C)=1
(Y*Y)=1/(A/C)
Y=sqrt(1/(A/C))
I did it again!

 

[mfb]

I moved the thread to our homework section.

Note that -1.3 is a solution as well.

 

[1plus1is10]

mfb, this wasn't homework. LOL

Mark, my first instinct regarding pow was right:
I discovered that sqrt(x) is the same as pow(x, 1/2)
I mention this because last night while I was trying to sleep it came to me.
I thought... What if I need 2 or 3 or 4 numbers between the 2 I have (e.g. A,1,2,C or A,1,2,3,C or A,1,2,3,4,C etc.)?
Well, after a while, it clicked with me what the answer is: pow(x, 1/3) or pow(x, 1/4) or pow(x, 1/5) etc.

Thanks again though, because I wrote about "equate two equations" (a.k.a. "substitution method") in my personal Math Notes.
Very helpful - I never know when I will need it again.

 

[Mark44]

mfb, this wasn't homework. LOL

But it's homework-like or could be a problem in a textbook, which means that the place to post it is here in the Homework & Coursework sections.

Mark, my first instinct regarding pow was right:
I discovered that sqrt(x) is the same as pow(x, 1/2)

pow() is a standard library function in C and C++ and possibly a few other languages. If all you need is the square root, most languages have a library function called sqrt(). However, why are you bringing in programming functions to what is a fairly simple problem in algebra?

For this problem, neither would have been helpful. In solving for b, the next to last step is $b^2 = 169 * 100 = 16,900$. If you naively take the square root (or the 1/2 power), you get b = 130, but b = -130 is also a solution of the equation $b^2 = 16,900$. This means that $y = \pm 1.3$. That's what @mfb was talking about at the bottom of post #6.

I mention this because last night while I was trying to sleep it came to me.
I thought... What if I need 2 or 3 or 4 numbers between the 2 I have (e.g. A,1,2,C or A,1,2,3,C or A,1,2,3,4,C etc.)?
Well, after a while, it clicked with me what the answer is: pow(x, 1/3) or pow(x, 1/4) or pow(x, 1/5) etc.

I have no idea what you're trying to do here.

Thanks again though, because I wrote about "equate two equations" (a.k.a. "substitution method") in my personal Math Notes.
Very helpful - I never know when I will need it again.

You aren't really "equating two equations" -- you're equating two expressions that have the same value. It doesn't make sense to set two equations equal to each other.

 

[1plus1is10]

I have no idea what you're trying to do here.

I'm sorry you have gotten frustrated with me.

As for what I'm doing... Well to do my best at using math words: I'm "interpolating a power series" (sort of).
And yes, it's computer code - C++ in fact... I create charts - I'm visual.
Numbers, formulas, "Math", ugh! - I know it's right when I "see" it.
I wish I understood it more, but I don't - I accept that (I don't let it get me down).
I've got you, right?

(we're almost neighbors. I'm up here in Bellingham - hello there. My wife and I are heading down to Bellevue tomorrow to watch a movie - 11yr anniversary)

 

[Mark44]

we're almost neighbors. I'm up here in Bellingham - hello there

Hello, back. I'm a couple of counties south of you.