logarithm help

**llkkjj24** · January 1st 2010, 11:57 AM

1.express in terms of log a, log b.

$\log \frac{1}{ab^4}$

2.express as a single logarithm.

$\frac{1}{2}\log 80 - \frac{1}{2}\log 5$

thanks

**Defunkt** · January 1st 2010, 12:12 PM

Originally Posted by llkkjj24

1.express in terms of log a, log b.

$\log \frac{1}{ab^4}$

2.express as a single logarithm.

$\frac{1}{2}\log 80 - \frac{1}{2}\log 5$

thanks

Use these two facts:

$log(x\cdot y) = log(x) + log(y)$

$log(\frac{x}{y}) = log(x) - log(y)$

And for the second, note that $\frac{1}{2}log(80) - \frac{1}{2}log(5) = \frac{1}{2}(log(80)-log(5))$ .

Can you do it now?

**llkkjj24** · January 1st 2010, 12:20 PM

ye, i did it before, but my answer didn't match of the book.
so was hoping for someone to work it out so i can see my mistakes

**pickslides** · January 1st 2010, 12:29 PM

Originally Posted by llkkjj24

so was hoping for someone to work it out so i can see my mistakes

Post your work so we can see your mistakes, it will be easier to assist you this way.

**llkkjj24** · January 2nd 2010, 05:29 AM

for 1 i had trouble of how to manipulate the fraction $\frac{1}{ab^4}$ .
is this the correct form? $\log a - \log b ^ -4$

for 2 i did using the laws of logs.

$\log 80^\frac{1}{2} - \log 5^\frac{1}{2}$

then $\log \frac{\log \sqrt{80}}{\log \sqrt{5}}$

**Stroodle** · January 2nd 2010, 06:16 AM

Using the properties posted above by Defunkt, you get:

$log\left ( \frac{1}{ab^4}\right )$

$\Rightarrow log(1)-\left[log(a)+log(b^4)\right]$

$\Rightarrow 0-log(a)-4log(b)$

$\Rightarrow -log(a)-4log(b)$

and for the second one:

$\frac{1}{2}log(80)-\frac{1}{2}log(5)$

$\Rightarrow \frac{1}{2}\left[log(80)-log(5)\right]$

$\Rightarrow\frac{1}{2}log(\frac{80}{5})$

$\Rightarrow\frac{1}{2}log(16)$

$\Rightarrow\log(16^\frac{1}{2})$

$\Rightarrow\log(4)$

**Stroodle** · January 2nd 2010, 06:27 AM

Originally Posted by llkkjj24

for 1 i had trouble of how to manipulate the fraction $\frac{1}{ab^4}$ .
is this the correct form? $\log a - \log b ^ -4$

for 2 i did using the laws of logs.

$\log 80^\frac{1}{2} - \log 5^\frac{1}{2}$

then $\log \frac{\log \sqrt{80}}{\log \sqrt{5}}$

You can also solve the second problem this way:

$\frac{1}{2}log(80)-\frac{1}{2}log(5)$

$\Rightarrow log\left ( \frac{\sqrt{80}}{\sqrt{5}}\right )$

$\Rightarrow log(4)$

(after rationalising and simplifying the fraction)

**llkkjj24** · January 2nd 2010, 08:26 AM

thanks , problem solved

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