1. ## Questions on logarithms

solving P and Q where they are integers

(25/16)^ -3/2 = 2^P * 5^Q

2. ## Re: Questions on logarithms

Originally Posted by Britt00
solving P and Q where they are integers
(25/16)^ -3/2 = 2^P * 5^Q
${\left( {\dfrac{{25}}{{16}}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( {\dfrac{{{5^2}}}{{{2^4}}}} \right)^{\dfrac{{ - 3}}{2}}} = \dfrac{{{5^{ - 3}}}}{{{2^{ - 6}}}} = ?$

3. ## Re: Questions on logarithms

no need for logs, note ...

$\left(\dfrac{25}{16}\right)^{-3/2} = \left(\dfrac{16}{25}\right)^{3/2} = \left(\dfrac{4^2}{5^2}\right)^{3/2} = \left(\dfrac{4}{5}\right)^3 = 4^3 \cdot 5^{-3}$

can you finish finding the values of P and Q from here?

4. ## Re: Questions on logarithms

Is it p=6 and q=-3 ?

5. ## Re: Questions on logarithms

Originally Posted by Britt00
Is it p=6 and q=-3 ?
Yes.

6. ## Re: Questions on logarithms

Originally Posted by Britt00
Is it p=6 and q=-3 ?
Substitute back into the original equation to check on your own.

$\left ( \dfrac{25}{16} \right )^{-(3/2)} = \left ( \dfrac{16}{25} \right )^{(3/2)} = \left ( \sqrt{\dfrac{16}{25} } \right )^3 =$

$\left ( \dfrac{4}{5} \right )^3 = \left ( \dfrac{2^2}{5} \right )^3 = \dfrac{2^6}{5^3} = 2^6 * 5^{-3}.$