# Thread: Power Law for Logarithms

1. ## Power Law for Logarithms

Solve for x in each of the following:

$\displaystyle log_2x = 3log_24$

--

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

Evaluate each of the following:

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

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

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

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

$\displaystyle log_2x = 3log_24$

--

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

Yes.

Originally Posted by Macleef
Evaluate each of the following:

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

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

Can you please show me how to solve these logs, I don't know where to start
We know that
$\displaystyle log_b(a^x) = x \cdot log_b(a)$

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

You do the second one.

-Dan

3. For the second equation, is the following what you do...

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

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

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

$\displaystyle = \frac {2}{3}$

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

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

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

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

$\displaystyle = \frac {2}{3}$
No, here we have another rule coming into play as well:
$\displaystyle log_b(xy) = log_b(x) + log_b(y)$

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

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

Can you finish from here?

-Dan

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

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

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

Can you finish from here?

-Dan

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

My work:

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

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

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

$\displaystyle = \frac {7}{3}$

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

My work:

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

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

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

$\displaystyle = \frac {7}{3}$
Yup!

-Dan