# Power Law for Logarithms

• Dec 9th 2007, 05:30 PM
Macleef
Power Law for Logarithms
Solve for x in each of the following:

$log_2x = 3log_24$

--

$logx = \frac{1}{3}log8$

Evaluate each of the following:

1) $log_2 \sqrt [5] {16}$

2) $log_525 \sqrt [3] {5}$

Can you please show me how to solve these logs, I don't know where to start
• Dec 9th 2007, 05:42 PM
topsquark
Quote:

Originally Posted by Macleef
Solve for x in each of the following:

$log_2x = 3log_24$

--

$logx = \frac{1}{3}log8$

Yes. :)

Quote:

Originally Posted by Macleef
Evaluate each of the following:

1) $log_2 \sqrt [5] {16}$

2) $log_525 \sqrt [3] {5}$

Can you please show me how to solve these logs, I don't know where to start

We know that
$log_b(a^x) = x \cdot log_b(a)$

So for the first one:
$log_2 ( \sqrt[5]{16} ) = \frac{1}{5} \cdot log_2(16) = \frac{1}{5} \cdot 4 = \frac{4}{5}$

You do the second one.

-Dan
• Dec 9th 2007, 05:51 PM
Macleef
For the second equation, is the following what you do...

$= \frac {1}{3} (log_525(5))$

$= \frac {1}{3} (log_5 5^{2} (5^{1}))$

$= \frac {1}{3} (2) (1)$

$= \frac {2}{3}$
• Dec 9th 2007, 06:07 PM
topsquark
Quote:

Originally Posted by Macleef
For the second equation, is the following what you do...

$= \frac {1}{3} (log_525(5))$

$= \frac {1}{3} (log_5 5^{2} (5^{1}))$

$= \frac {1}{3} (2) (1)$

$= \frac {2}{3}$

No, here we have another rule coming into play as well:
$log_b(xy) = log_b(x) + log_b(y)$

So....
$log_5(25\sqrt[3]{5})$

$= log_5(25) + log_5(\sqrt[3]{5})$

Can you finish from here?

-Dan
• Dec 9th 2007, 06:17 PM
Macleef
Quote:

Originally Posted by topsquark
No, here we have another rule coming into play as well:
$log_b(xy) = log_b(x) + log_b(y)$

So....
$log_5(25\sqrt[3]{5})$

$= log_5(25) + log_5(\sqrt[3]{5})$

Can you finish from here?

-Dan

Is the answer $\frac {7}{3}$?

My work:

$= log_55^{2} + \frac {1}{3} log_55$

$= 2(log_55) + \frac {1}{3} log_55$

$= 2 + \frac {1}{3}$

$= \frac {7}{3}$
• Dec 9th 2007, 06:23 PM
topsquark
Quote:

Originally Posted by Macleef
Is the answer $\frac {7}{3}$?

My work:

$= log_55^{2} + \frac {1}{3} log_55$

$= 2(log_55) + \frac {1}{3} log_55$

$= 2 + \frac {1}{3}$

$= \frac {7}{3}$

Yup! (Handshake)

-Dan