### Logarithms Part I

Intro to logarithms

$ \begin{aligned}

y &= a^x \\

log_{a} y &= log_{a} a^x \\

log_{a} y &= x\\

\end{aligned}$

y, $\ a^x $ is known as the argument

a is known as the base

x is known as the exponent or the power

It is important to recognize that the base of the argument (a) equals to the base of the log (a)

Example 1

$ \begin{aligned}

y &= a^x \\

3 &= 10^x\\

log_{10} 3 &= log_{10} 10^x \\

log_{10} 3 &= x\\

\end{aligned}$

Example 2

$ \begin{aligned}

5 &= e^x\\

log_{e} 5 &= log_{e} e^x \\

log_{e} 5 &= x\\

x &= 1.61 \quad \text{to 3 s.f.}

\end{aligned}$

The next example will show you how Taking log on both sides can help you to answer questions

Example 3

Solve $ 10^{2x} - 2 ( 10^{x} ) -3 =0 $

We start by using the Substitution $ A = 10^x $

$\begin{aligned}

A^2 - 2A - 3 = 0 \\

(A - 3) ( A + 1) = 0 \\

A = 3 \qquad A = -1 \\

\text{Recall that } A = 10^x \\

10^x = 3 \quad 10^x = -1 \\

10^x \neq -1 \\

\text{hence we reject } 10^x = -1 \\

10^x = 3 \\

\text{ taking log on both sides } \\

\log_{10} 10^x = \log_{10} 3 \\

\text{ use calculator to evaluate } \log_{10} 3 \\

x = 0.477 \\

\end{aligned} $

Logarithms Part II will illustrate how we can use the "Taking log on both sides" technique in practical situations

$ \begin{aligned}

y &= a^x \\

log_{a} y &= log_{a} a^x \\

log_{a} y &= x\\

\end{aligned}$

y, $\ a^x $ is known as the argument

a is known as the base

x is known as the exponent or the power

It is important to recognize that the base of the argument (a) equals to the base of the log (a)

Example 1

$ \begin{aligned}

y &= a^x \\

3 &= 10^x\\

log_{10} 3 &= log_{10} 10^x \\

log_{10} 3 &= x\\

\end{aligned}$

- log base 10 is commonly used as it can be evaluated using a calculator.

- In most cases, it is abbreviated as lg. $\ log_{10} $ is equivalent to $\ lg $.

- Another number that is commonly used is 2.718, commonly known as e.

- log base e (2.718) can also be evaluated using a calculator.

- $\ log_{e} $ is equivalent to $\ ln $.

Example 2

$ \begin{aligned}

5 &= e^x\\

log_{e} 5 &= log_{e} e^x \\

log_{e} 5 &= x\\

x &= 1.61 \quad \text{to 3 s.f.}

\end{aligned}$

The next example will show you how Taking log on both sides can help you to answer questions

Example 3

Solve $ 10^{2x} - 2 ( 10^{x} ) -3 =0 $

We start by using the Substitution $ A = 10^x $

$\begin{aligned}

A^2 - 2A - 3 = 0 \\

(A - 3) ( A + 1) = 0 \\

A = 3 \qquad A = -1 \\

\text{Recall that } A = 10^x \\

10^x = 3 \quad 10^x = -1 \\

10^x \neq -1 \\

\text{hence we reject } 10^x = -1 \\

10^x = 3 \\

\text{ taking log on both sides } \\

\log_{10} 10^x = \log_{10} 3 \\

\text{ use calculator to evaluate } \log_{10} 3 \\

x = 0.477 \\

\end{aligned} $

Logarithms Part II will illustrate how we can use the "Taking log on both sides" technique in practical situations

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