Solving logarithmic functions.

**mankvill** · July 20th 2008, 07:31 PM

It escapes me how to do these at the moment.

Time is a factor!

<3

**fobos3** · July 20th 2008, 10:22 PM

$3 ln(x^3 y^4) +5 ln(x^{-1})-4 ln(y)$
$3 ln(x^3)+3ln(y^4)-5ln(x)-4ln(y)$
$9ln(x)+12ln(y)-5ln(x)-4ln(y)$
$4ln(x)+8ln(y)$

I've used the following properties:
$ln(ab)=ln(a)+ln(b)$
$ln(a^n)=n*ln(a)$
$\frac {1}{x}=x^{-1}$

**tutor** · July 21st 2008, 05:42 AM

Hi,

=log(x^9y^12)+log(1/x^5)-log(y^4)

=log(x^9y^12)+log(1/x^5.y^4)

=log(x^9y^12.1/x^5.y^4)

=logx^4y^8

here we used the formula for logab=loga+logb

log(a/b)=loga-logb

**Soroban** · July 21st 2008, 07:45 AM

Hello, mankvill!

I'll give it a try . . .

$3\!\cdot\!\log(x^3y^4) + 5\!\cdot\!\log\left(\frac{1}{x}\right) - 4\!\cdot\!\log(y)$

We have: . $\log(x^3y^4)^3 + \log\left(\frac{1}{x}\right)^5 - \log(y)^4$

. . . . . . $= \;\underbrace{\log(x^9y^{12}) + \log\left(\frac{1}{x^5}\right)} - \log(y^4)$

. . . . . . $= \quad\;\;\overbrace{\log\left(x^9y^{12}\cdot\frac{ 1}{x^5}\right)} - \log(y^4)$

. . . . . . $= \;\qquad\quad\log(x^4y^{12}) - \log(y^4)$

. . . . . . $= \;\qquad\qquad\log\left(\frac{x^4y^{12}}{y^4}\righ t)$

. . . . . . $= \qquad\qquad\;\; \log(x^4y^8)$

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