Logarithms  (Page 2/3)

Laws of logarithms

Just as for the exponents, logarithms have some laws which make working with them easier. These laws are based on the exponential laws and are summarised first and then explained in detail.

Logarithm law 1: ${log}_{a}1=0$

For example,

and

Logarithm law 1: ${log}_{a}1=0$ :

Simplify the following:

1. ${log}_2\left(1\right)+5$
2. ${log}_{10}\left(1\right)\times 100$
3. $3\times {log}_{16}\left(1\right)$
4. ${log}_x\left(1\right)+2xy$
5. $\frac{{log}_y\left(1\right)}{x}$

Logarithm law 2: ${log}_a\left(a\right)=1$

For example,

and

Logarithm law 2: ${log}_a\left(a\right)=1$ :

Simplify the following:

1. ${log}_2\left(2\right)+5$
2. ${log}_{10}\left(10\right)\times 100$
3. $3\times {log}_{16}\left(16\right)$
4. ${log}_x\left(x\right)+2xy$
5. $\frac{{log}_y\left(y\right)}{x}$

When the base is 10, we do not need to state it. From the work done up to now, it is also useful to summarise the following facts:

1. $log1=0$
2. $log10=1$
3. $log100=2$
4. $log1000=3$

Logarithm law 3: ${log}_a(x·y)={log}_a\left(x\right)+{log}_a\left(y\right)$

The derivation of this law is a bit trickier than the first two. Firstly, we need to relate $x$ and $y$ to the base $a$ . So, assume that $x=a^m$ and $y=a^n$ . Then from Equation  [link] , we have that:

This means that we can write:

For example, show that $log(10·100)=log10+log100$ . Start with calculating the left hand side:

The right hand side:

Both sides are equal. Therefore, $log(10·100)=log10+log100$ .

Logarithm law 3: ${log}_a(x·y)={log}_a\left(x\right)+{log}_a\left(y\right)$ :

Write as seperate logs:

1. ${log}_2(8\times 4)$
2. ${log}_8(10\times 10)$
3. ${log}_{16}\left(xy\right)$
4. ${log}_z\left(2xy\right)$
5. ${log}_x\left(y^2\right)$

Logarithm law 4: ${log}_a\left(\frac{x}{y}\right)={log}_a\left(x\right)-{log}_a\left(y\right)$

The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.

For example, show that $log\left(\frac{10}{100}\right)=log10-log100$ . Start with calculating the left hand side:

The right hand side:

Both sides are equal. Therefore, $log\left(\frac{10}{100}\right)=log10-log100$ .

Logarithm law 4: ${log}_a\left(\frac{x}{y}\right)={log}_a\left(x\right)-{log}_a\left(y\right)$ :

Write as seperate logs:

1. ${log}_2\left(\frac{8}{5}\right)$
2. ${log}_8\left(\frac{100}{3}\right)$
3. ${log}_{16}\left(\frac{x}{y}\right)$
4. ${log}_z\left(\frac{2}{y}\right)$
5. ${log}_x\left(\frac{y}{2}\right)$

Logarithm law 5: ${log}_a\left(x^b\right)=b{log}_a\left(x\right)$

Once again, we need to relate $x$ to the base $a$ . So, we let $x=a^m$ . Then,

For example, we can show that ${log}_2\left(5^3\right)=3{log}_2\left(5\right)$ .

Therefore, ${log}_2\left(5^3\right)=3{log}_2\left(5\right)$ .

Logarithm law 5: ${log}_a\left(x^b\right)=b{log}_a\left(x\right)$ :

Simplify the following:

1. ${log}_2\left(8^4\right)$
2. ${log}_8\left({10}^{10}\right)$
3. ${log}_{16}\left(x^y\right)$
4. ${log}_z\left(y^x\right)$
5. ${log}_x\left(y^{2x}\right)$

Logarithm law 6: ${log}_a\left(\sqrt[b]{x}\right)=\frac{{log}_a\left(x\right)}{b}$

The derivation of this law is identical to the derivation of Logarithm Law 5 and is left as an exercise.

For example, we can show that ${log}_2\left(\sqrt[3]{5}\right)=\frac{{log}_{2}5}{3}$ .

Therefore, ${log}_2\left(\sqrt[3]{5}\right)=\frac{{log}_{2}5}{3}$ .

Logarithm law 6: ${log}_a\left(\sqrt[b]{x}\right)=\frac{{log}_a\left(x\right)}{b}$ :

Simplify the following:

1. ${log}_2\left(\sqrt[4]{8}\right)$
2. ${log}_8\left(\sqrt[10]{10}\right)$
3. ${log}_{16}\left(\sqrt[y]{x}\right)$
4. ${log}_z\left(\sqrt[x]{y}\right)$
5. ${log}_x\left(\sqrt[2x]{y}\right)$