# Expanding Log problem...

• Oct 13th 2009, 02:34 PM
Savior_Self
Expanding Log problem...
Expand as the sum of individual logarithms, each of whose argument is linear: $\displaystyle log(\frac{xy^2}{z^4})$

• Oct 13th 2009, 03:56 PM
Defunkt
Quote:

Originally Posted by Savior_Self
Expand as the sum of individual logarithms, each of whose argument is linear: $\displaystyle log(\frac{xy^2}{z^4})$

Using the following logarithm rules:

$\displaystyle log(ab) = log(a) + log(b)$
$\displaystyle log(\frac{a}{b})= log(a) - log(b)$
$\displaystyle log(a^n) = n \cdot log(a)$

I will start, see if you can finish:

$\displaystyle log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$

...
• Oct 13th 2009, 04:05 PM
Savior_Self
Quote:

Originally Posted by Defunkt
Using the following logarithm rules:

$\displaystyle log(ab) = log(a) + log(b)$
$\displaystyle log(\frac{a}{b})= log(a) - log(b)$
$\displaystyle log(a^n) = n \cdot log(a)$

I will start, see if you can finish:

$\displaystyle log(\frac{xy^2}{z^4}) = log(xy^2) - log(z^4) = log(xy^2) - 4log(z)$

...

so...

$\displaystyle log(x) + log(y^2) - 4log(z) = log(x) + 2log(y) - 4log(z)$

$\displaystyle log(x) + 2log(y) - 4log(z)$

Look good?
• Oct 13th 2009, 04:50 PM
Defunkt
Quote:

Originally Posted by Savior_Self
so...

$\displaystyle log(x) + log(y^2) - 4log(z) = log(x) + 2log(y) - 4log(z)$

$\displaystyle log(x) + 2log(y) - 4log(z)$