Elementary Functions Logarithm question

I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

log (x^2)*(y^3) = 2*log x*(y^3)

2*3*log x*y

6*(log x + log y)

I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.

Re: Elementary Functions Logarithm question

Consider $\displaystyle \log (ab) = \log a + \log b$

and $\displaystyle \log a^b = b\log a$

threrefore

$\displaystyle \displaystyle \log (x^2y^3) = \log (x^2)+ \log (y^3) = 2\log x+ 3\log y$

Re: Elementary Functions Logarithm question

Quote:

Originally Posted by

**bkbowser** I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

log (x^2)*(y^3) = 2*log x*(y^3)

2*3*log x*y

6*(log x + log y)

I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.

$\displaystyle x^2= x*x$, $\displaystyle y^3= y*y*y$ so $\displaystyle x^2y^3$ would be x*x*y*y*y. $\displaystyle 6(log x+ log y)= 6log(xy)=log(x^6y^6)$ and $\displaystyle x^6y^6= x*x*x*x*x*x*y*y*y*y*y*y$

Re: Elementary Functions Logarithm question

Hello, bkbowser!

You are taking liberties with Algebra . . .

Quote:

$\displaystyle \text{I'd lik\,\!e to know why }\,\log (x^2y^3) \:=\:2\log x + 3\log y$

. . $\displaystyle \text{and not }6(\log x + \log y).$

$\displaystyle \log (x^2y^3) \:=\: 2\log (xy^3)$ .This is wrong!

Suppose your answer is correct . . .

You have: .$\displaystyle 6(\log x + \log y) \;=\; 6\log(xy) \;=\;\log(xy)^6 \;=\;\log(x^6y^6) $

. . See?