# Properties of the product and the quotient of the logarithms

Let's learn how to create and solve exercises based on the properties of the product and the quotient of the logarithms.

A property of the logarithms is: $$log_a (x\cdot y)=log_a x+log_a y$$$Or, in other words, the logarithm of the product of two numbers is the sum of the logarithms of these numbers. $$log_2 (8\cdot64)=log_2 8+log_2 64=log_2 2^3+log_2 2^6=3+6=9$$ Also: $$log_5 (125\cdot625)=log_5 125+log_5 625=log_5 5^3+log_5 5^4=3+4=7$$ When we have a product of logarithms we have to add, but when we have a quotient we have to subtract, therefore the second property of the logarithms is: $$log_a \dfrac{x}{y}=log_a x - log_a y$$$

Or, in other words, the logarithm of the quotient of two numbers is equal to the logarithm of the numerator minus the logarithm of the denominator.

$$log_3 \dfrac{9}{81}=log_3 9-log_3 81= log_3 3^2-log_3 3^4=2-4=-2$$

Also:

$$log_{10} \dfrac{0,001}{0,01}=log_{10} 0,001-log_{10} 0,01= log_{10} 10^{-3}-log_{10} 10^{-2}=-3+2=-1$$

The base of the last example $$(10)$$ is very common. In fact, it is one of two types of logarithms that scientific pocket calculators calculate. It is called decimal logarithm and it is usually abbreviated as log, without any need to specify the base.

Another type of logarithm, very common as well, is the natural or neperian, which has number $$e$$ as its base and is abbreviated $$ln$$.

The properties of the product and the quotient of the logarithms can be used together in order to simplify expressions.

It is possible to group the following expression into just one logarithm: $$ln127-ln481-ln7+ln74$$$First, the elements with the same sign are grouped: $$(ln127+ln74)-(ln481+ln7)$$$ Applying the properties of the product and the quotient, we obtain: $$(ln127+ln74)-(ln481+ln7)=ln(\dfrac{127\cdot74}{481\cdot7})$$$The following expression sums up the properties of the logarithms seen until now. We have to try to reduce the expression in just one logarithm: $$log10^{-2}-log\dfrac{3}{4}+log13^{\frac{1}{3}}-log5$$$ First we group the elements with the same sign: $$\Big(log10^{-2}+log13^{\frac{1}{3}}\Big)-\Big(log\dfrac{3}{4}+log5\Big)$$$Now we apply the properties of the product, quotient and the power of a logarithm: $$\Big(log10^{-2}+log13^{\frac{1}{3}}\Big)-\Big(log\dfrac{3}{4}+log5\Big)=log\dfrac{10^{-2}\cdot13^{\frac{1}{3}}}{\dfrac{3}{4}\cdot5}=$$$ $$=-2\cdot\dfrac{1}{3}log\dfrac{1\cdot13}{\dfrac{15}{4}}=-\dfrac{2}{3}log\dfrac{52}{15}$$$Notice that the product and quotient properties of the logarithms come from the corresponding properties of the powers: $$log_a (x\cdot y)=log_a x+log_a y$$$ Because $$a^n \cdot a^m=a^{n+m}$$, and applying logarithms implies calculating $$n$$ and $$m$$.

On the other hand: $$log_a \dfrac{x}{y}=log_a x-log_a y$$\$ since $$\dfrac{a^n}{a^m}=a^{n-m}$$