Elementary Functions Logarithm question

**bkbowser** · June 29th 2011, 02:15 PM

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.

**pickslides** · June 29th 2011, 02:25 PM

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

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

threrefore

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

**HallsofIvy** · June 29th 2011, 02:57 PM

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.

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

**Soroban** · June 29th 2011, 04:54 PM

Hello, bkbowser!

You are taking liberties with Algebra . . .

$\text{I'd lik\,\!e to know why }\,\log (x^2y^3) \:=\:2\log x + 3\log y$
. . $\text{and not }6(\log x + \log y).$

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

Suppose your answer is correct . . .

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

. . See?

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