How do I rearrange this for y?

log(10)y=3log(10)x+2

thx

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- Jul 11th 2013, 07:23 PMJellyOnionRearranging log equation 2
How do I rearrange this for y?

log(10)y=3log(10)x+2

thx - Jul 11th 2013, 08:02 PMibduttRe: Rearranging log equation 2
In fact what you want is not clear

It may be written as

log 10y - 3 lof 10x = 2

log (10y)/( 1000 x ^3) = 2

If the log is to the base 10 then we have

(10y)/( 1000 x ^3) = 10 ^ 2 etc please be specific with your question so as to enable us to help and avoid guess work - Jul 11th 2013, 11:27 PMJellyOnionRe: Rearranging log equation 2
Sorry I didn't realise.

I did mean base 10 thx - Jul 12th 2013, 07:18 AMithanareshbabuRe: Rearranging log equation 2
log(y/x^3)=10^2

- Jul 12th 2013, 07:38 AMtopsquarkRe: Rearranging log equation 2
Recall that $\displaystyle y= 10^{log(y)}$. Does this give you any ideas how to simplify the LHS?

-Dan - Jul 12th 2013, 09:10 AMSorobanRe: Rearranging log equation 2
Hello, JellyOnion!

Since the base is 10 (ten), we can drop the subscript.

Quote:

$\displaystyle \text{Solve for }y\!:\;\log y \:=\:3\log x+2$

We have: .$\displaystyle \log y \;=\; 3\log x + 2\cdot 1 \;=\;3\log x + 2\log(10) $

. . . . . . . . $\displaystyle \log y \;=\;\log(x^3) + \log(10^2) \;=\;\log(x^3) + \log(100)$

. . . . . . . . $\displaystyle \log y \;=\;\log(100x^3)$

Therefore:. . .$\displaystyle y \;=\;100x^3$

- Jul 12th 2013, 09:34 AMHallsofIvyRe: Rearranging log equation 2
Or log(y)= 3log(x)+ 2 so log(y)- 3 log(x)= log(y/x^3)= 2

y/x^3= 100, y= 100x^3 - Jul 13th 2013, 06:07 PMJellyOnionRe: Rearranging log equation 2
Thank u everyone, much appreaciated