Simplifying Logarithm

Question

I am experiencing difficulty in simplifying the following equation:

$$\log_{2}\frac 8{125} \;- 3\log_{2}\frac 3{5} \; -4\log_{2}\frac 1{2}.$$

Answer 1

The basic rules of logs

$\log ab = \log a + \log b\\ \log \frac {a}{b} = \log a - \log b\\ \log a^n = n\log a$

They are really all the same rule... just presented differently..

To simplify, you will need to apply that second line to break up the fractions.

$\log 2^3 -\log 125 - 3\log 3+3\log 5 - 4 \log 1 +4\log 2$

$8 = 2^3$ and $125 = 5^3$ use the 3rd rule to simplify that, and the second one will cancel with the $3\log 5$ term.

$3\log 2 - 3\log 3 - 4 \log 1 +4\log 2$

What else $\log_2 2 = 1$ and $\log 1 = 0$

$7-3\log_2 3$

Answer 2

Hint

$$\log ab=\log a+\log b$$ $$\log \frac{a}{b}=\log a-\log b$$ $$k\log a=\log a^k$$

Answer 3

HINTS

$$\log_{2}\frac 8{125}=\log_{2}\frac {2^3}{5^3}=\log_2 2^3 - \log_25^3$$

$$3\log_{2}\frac 3{5}=3(\log_23 \,- \log_25) $$

$$4\log_{2}\frac 1{2}=-4\log_22$$