# How do you write the following expression as a single logarithm: log 5 - log x - log y?

$\log \left(\frac{5}{x y}\right)$

#### Explanation:

We have two properties in Logarithms that can be used in solving this question

• ${\log}_{m} \left(a\right) + {\log}_{m} \left(b\right) = {\log}_{m} \left(a \cdot b\right)$

proof :-

Let :

${\log}_{m} \left(a\right) = p \implies {m}^{p} = a$ ------1

and

${\log}_{m} \left(b\right) = q \implies {m}^{q} = b$ ------2

So,

From 1 and 2 ;

$a \cdot b = {m}^{p} \cdot {m}^{q}$

$\therefore a \cdot b = {m}^{p + q}$

By writing $\uparrow$ in logarithmic form

${\log}_{m} \left(a b\right) = p + q$

$\implies {\log}_{m} \left(a b\right) = {\log}_{m} \left(a\right) + {\log}_{m} \left(b\right)$

and

• ${\log}_{m} \left(a\right) - {\log}_{m} \left(b\right) = {\log}_{m} \left(\frac{a}{b}\right)$

proof:-

Let :

${\log}_{m} \left(a\right) = p \implies {m}^{p} = a$ -------1

and

${\log}_{m} \left(b\right) = q \implies {m}^{q} = b$ ------2

So,

From 1 and 2 ;

$\frac{a}{b} = {m}^{p} / {m}^{q}$

$\therefore \frac{a}{b} = {m}^{p - q}$

By writing $\uparrow$ in logarithmic form

${\log}_{m} \left(\frac{a}{b}\right) = p - q$

$\implies {\log}_{m} \left(\frac{a}{b}\right) = {\log}_{m} \left(a\right) - {\log}_{m} \left(b\right)$

The given question is :

$\log \left(5\right) - \log \left(x\right) - \log \left(y\right)$

$\implies \log \left(5\right) - \left(\log \left(x\right) + \log \left(y\right)\right)$

$\implies \log \left(5\right) - \left(\log \left(x \cdot y\right)\right)$

$\implies \log \left(\frac{5}{x \cdot y}\right)$

$\implies \log \left(\frac{5}{x y}\right)$