rewriting a formula

• Dec 4th 2007, 03:01 PM
Raven
rewriting a formula
Here's the original formula.

B = log (X/Y)

I have to rewrite it to be Y = (something). I tried to do it but I'm not real sure I did it right.

I thought I could unlog it first

10^B = X/Y

Then divide by Y to get Y on the other side.

10^B / Y = X

Then I turned it over.

Y / 10^B = 1 / X

Then divide both sides by 10^B.

Y = 1 / ( X / 10^B )

Am I doing it right? Or close to right?
Raven
• Dec 4th 2007, 03:04 PM
Jhevon
Quote:

Originally Posted by Raven
...

Y / 10^B = 1 / X

Then divide both sides by 10^B.

Y = 1 / ( X / 10^B )

Am I doing it right? Or close to right?
Raven

isn't the 10^B already dividing the left? we need to multiply both sides by 10^B
• Dec 4th 2007, 03:17 PM
Raven
So from

Y / 10^B = 1 / X

it's now

Y = (10^B)(1/X)

and from that

Y = 10^B / X

ok?
Raven
• Dec 4th 2007, 03:19 PM
Jhevon
Quote:

Originally Posted by Raven
So from

Y / 10^B = 1 / X

it's now

Y = (10^B)(1/X)

and from that

Y = 10^B / X

ok?
Raven

nope.

you made the same error with Y earlier, i just didn't catch it until now. you should have multiplied both sides by Y when you divided
• Dec 4th 2007, 03:31 PM
Raven
ok, do over

B = log (X/Y)
10^B = X/Y
(10^B)Y = X
Y = X / 10^B
• Dec 4th 2007, 03:34 PM
Jhevon
Quote:

Originally Posted by Raven
ok, do over

B = log (X/Y)
10^B = X/Y
(10^B)Y = X
Y = X / 10^B

yes, correct (Clapping)

An alternate solution:

$\displaystyle B = \log \left( \frac XY \right)$

$\displaystyle \Rightarrow B = \log X - \log Y$

$\displaystyle \Rightarrow \log Y = \log X - B$

$\displaystyle \Rightarrow Y = 10^{\log X - B}$

$\displaystyle \Rightarrow Y = 10^{\log X} \cdot 10^{-B}$

$\displaystyle \Rightarrow Y = 10^{-B}X$