**Motivation**

When we were dealing with exponentiation, we wondered, for example, what is the result of $2^{4}.$ We concluded that the exponent (in this case number 4) tells us how many times we need to multiply the base (in this case number 2) by itself. The result of that exponentiation is $16$. But, what if we want to know which number $x$ (exponent) produces: $2^{x}=16$?

For this purpose, we need to introduce **logarithms**.

If we have $2^{4}=16$, then the equivalent statement using logarithms is $log_{2}16=4$. So, general rule is:

$$log_{a}y=x \Leftrightarrow a^{x}=y,$$ where $a$ is **the base**, $x$ **the exponent **and $y$ **the result**. Therefore, $$a^{log_{a}y}=y.$$

Therefore, logarithm gives us the exponent for one number to become another number.

*Example 1: *

a) What is the $log_{3}27$?

b) What is the $log_{5}125$?

*Solution:*

a) This is equivalent to: $3^{x}=27$ and we know that $3^{3}=27$, so we conclude that $x=3$. Therefore, $log_{3}27=3$.

b) This is equivalent to: $5^{x}=125$ and we know that $5^{3}=125$, so we conclude that $x=3$. Therefore, $log_{5}125=3$.

**Common logarithm**

**A common logarithm **is any logarithm with **base 10**. It is also known as **decimal**, **decadic** or **Briggsian logarithm**. The common logarithm of number $x$ is written as: $logx$.

* Example 2: *What is the $log 100$?

** Solution: **This is equivalent to: $10^{x}=100$ and we know that $10^{2}=100$, so we conclude that $x=2$. Therefore, $log 100=2$.

**Natural logarithm**

**Natural logarithm** is any logarithm with **base $e$**, which is a irrational number whose value is approximately equal to $2.71828$. Precisely, $$e= \lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^x\approx 2.71828,$$ or $$e= \lim_{x\to 0} \left(1 + x\right)^\frac{1}{x}\approx 2.71828.$$

Number $e$ is also known as **Euler’s number** or **Napier’s constant**. The natural logarithm of number $x$

is written as: $lnx=log_{e}x$.

**Logarithmic and exponential function**

Logarithmic function $$f: \mathbf{R^{+}} \rightarrow \mathbf{R}, f(x)=log_{a}x, x \in \mathbf{R^{+}}, a >0, a \neq 1$$ is the **inverse function** of exponential function $$g: \mathbf{R} \rightarrow \mathbf{R^{+}}, g(x)=a^x, x \in \mathbf{R}, a >0, a \neq 1.$$

**Logarithms are defined only for positive numbers! **

**Example 3: **

a) What is the $log_{5} (-25)$?

b) What is the $log_{3} 0$?

**Solution:**

a) $log_{5} (-25)$ doesn’t exist, since -25 is negative number.

b) $log_{3} 0$ doesn’t exist, since logarithms are defined only for positive numbers.

The values of logarithms can be negative (because the codomain of logarithmic function is set $\mathbf{R}$), which we will see in the following example.

**Example 4: **

a) What is the $log_{7} \frac{1}{7}$?

b) What is the $log_{5}0.006$?

**Solution:**

a) $log_{7} \frac{1}{7} = -1$, since we know $7^{-1}= \frac{1}{7^{1}}=\frac{1}{7}$.

b) This is equivalent to: $5^x=0.006$. Obviously, the logarithm will be a negative number, since $0.006 < 5$. Therefore, we have to divide number $1$ by number $5$, $4$ times: $\frac{1}{5\cdot 5 \cdot 5 \cdot 5 }=\frac{1}{5^{4}}=5^{-4}=0.006$. We conclude that $x=log_{5}0.006=-4$.

**Properties of logarithms**

**Product rule: $log_{a}(x \cdot y)=log_{a}x + log_{a}y$**

**Quotient rule: $log_{a}\left(\frac{x}{y}\right)=log_{a}x – log_{a}y$**

**Power rule: $log_{a}x^n = n \cdot log_{a}x$**

**Base switch rule: $log_{a}b =\frac{1}{log_{b}a} $**

**Base change rule: $log_{a}x = \frac {log_{b}x}{log_{b}a}$**

**Logarithm of $1$: $log_{a}1 = 0$**

**Logarithm of the base: $log_{a}a=1$**

**If $a>1$ and $x_{1}<x_{2}$, then $log_{a} x_{1} < log_{a} x_{2}$. **

** If $0<a<1$ and $x_{1}<x_{2}$, then $log_{a} x_{1} > log_{a} x_{2}$.**

**If $log_{a} x_{1} = log_{a} x_{2}$, then $x_{1}=x_{2}$.**

**Derivative of logarithm: $f(x)=log_{a}x \Rightarrow f'(x)= \frac{1}{x lna}$**

* Example 5: *Compute the following logarithms.

a) $log_{4}(16 \cdot 256)$

b) $log_{3}\frac{27}{81}$

**Solution:**

a) Use the product rule to write the given logarithm as sum of logarithms:

$$log_{4}(16 \cdot 256) = log_{4}16 + log_{4}256 $$

If it is possible, simplify each addend: $$= log_{4}4^2 + log_{4}4^4$$

Use the power rule: $$= 2 log_{4}4 + 4 log_{4}4$$

Since logarithm of the base is $1$, we have $$=2 \cdot 1 + 4 \cdot 1 = 6$$

Therefore, $log_{4}(16 \cdot 256) = 6.$

b) Use the quotient rule to write the given logarithm as difference of logarithms:

$$log_{3}\frac{27}{81}=log_{3}27 – log_{3}81$$

If it is possible, simplify each addend: $$= log_{3} 3^3 – log_{3}3^4$$

Use the power rule: $$= 3 \cdot log_{3} 3 – 4 \cdot log_{3} 3$$

Since logarithm of the base is $1$, we have $$= 3 \cdot 1 – 4 \cdot 1 = -1.$$

Therefore, $log_{3}\frac{27}{81}= -1.$

* Example 6: *Use the logarithm rules to expand $log_{7} \frac{xy}{zw}$ into four simpler terms.

**Solution:**

We can rewrite this expression using the quotient rule:

$$log_{7} \frac{xy}{zw}= log_{7} xy – log_{7} zw.$$

Now we can use the product rule: $$= log_{7}x + log_{7}y – (log_{7}z + log_{7}w)$$

Finally, we have $$=log_{7}x + log_{7}y – log_{7}z – log_{7}w.$$