# How do you condense Log_4 (20) - Log_4 (45) + log_4 (144)?

May 27, 2018

$3$

#### Explanation:

$\text{using the "color(blue)"laws of logarithms}$

•color(white)(x)logx+logy=log(xy)

•color(white)(x)logx-logy=log(x/y)

•color(white)(x)log_b x=nhArrx=b^n

${\log}_{4} 20 + {\log}_{4} 144 - {\log}_{4} 45$

$= {\log}_{4} \left(\frac{20 \times 144}{45}\right)$

$= {\log}_{4} \left(64\right) = n$

$64 = {4}^{3} = {4}^{n} \Rightarrow n = 3$

May 27, 2018

${\log}_{4} \left(\frac{20}{45} \cdot 144\right) = {\log}_{4} \left(64\right) = 3$

#### Explanation:

The log product and quotient rules allow us to combine these terms, as long as the logs have the same base:
${\log}_{b} \left(x\right) + {\log}_{b} \left(y\right) = {\log}_{b} \left(x \cdot y\right)$
${\log}_{b} \left(x\right) - {\log}_{b} \left(y\right) = {\log}_{b} \left(\frac{x}{y}\right)$

All the terms in this problem have logs of base 4, so we can apply these rules:
${\log}_{4} \left(20\right) - {\log}_{4} \left(45\right) + {\log}_{4} \left(144\right)$
$= {\log}_{4} \left(\frac{20}{45} \cdot 144\right)$
$= {\log}_{4} \left(64\right)$

Of course, the result of a log is the exponent of the base that will give you that number: ${\log}_{b} \left(x\right) = y$ means that ${b}^{y} = x$
So, since ${4}^{3} = 64$, then
${\log}_{4} \left(64\right) = 3$