### Logarithms Part III

Laws of logarithm

Product Law

$\ \log_a mn = \log_a m + \log_a n $

note that the logarithms are of the same base

Quotient Law

$\ \log_a {\frac{m}{n}} = \log_a m - \log_a n $

note that the logarithms are of the same base

Power Law

$\ \log_a m^n = n\log m $

Important: $\ \log_a xy^n \ne n\log_a xy $

The power n belongs to the base y only. The power n does not belong to base x

$ \begin{aligned}

\log_a xy^n &= \log_a x + \log_a y^n \quad \text{ applying power law} \\

\log_a xy^n &= \log_a x + n\log_a y \\

\end{aligned}$

Using the Laws of logarithms to solve equations

Strategy

Example 1

$ \begin{aligned}

\lg {(5y +2)} &= 1 +\lg y \\

\lg {(5y +2)} &= \lg 10 +\lg y \qquad \text{Note that} 1 = \lg10 \\

\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\

5y + 2 &= 10y \qquad \text{(comparing arguments)} \\

5y &= 2 \\

y &= \frac{2}{5} \\

\end{aligned}$

Example 2

$ \begin{aligned}

\log_3 {(5y +7)} &= 2 +\log_3 {(y + 4)} \\

\log_3 {(5y +7)} &= 2\log_3{3} + \log_3 {(y + 4)} \qquad \text{Note that} 2 = 2\log_3{3} \\

\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\

5y + 2 &= 10y \qquad \text{(comparing arguments)} \\

5y &= 2 \\

y &= \frac{2}{5} \\

\end{aligned}$

Example 3

Solve the following equation

$\log(x+20) - \log(x+2) = \log x $

$\begin{aligned}

\log(x+20) - \log(x+2) &= \log x \\

\log \frac{ x+20}{ x+2 } &= \log x \\

\frac{ x+20}{ x+2 } &= x \\

x + 20 &= x^2 + 2x \\

x^2 + x - 20 &= 0 \\

(x + 5)( x - 4) &= 0 \\

x = -5 \; \text{rejected} \qquad x = 4 \\

\end{aligned} $

$ x = -5 $ is rejected as you cannot log a negative value

Example 4

Solve the following equation: $ 3^{x-4} = 5^{x-1} $

$\begin{aligned}

3^{x-4} &= 5^{x-1} \\

\log{3^{x-4}} &= \log{5^{x-1}} \qquad \text{log both sides} \\

(x-4) \log{3} &= (x-1) \log{5} \\

x( \log{3} - \log{5}) &= 4\log{3} - \log{5} \\

x &= \frac{4\log{3} - \log{5}}{ \log{3} - \log{5}} \\

\end{aligned} $

$ log_{10} 3 $ and $ log_{10} 5 $ can both be evaluated using the calculator

Take home message

To solve log equations:

What's Next?

Click here to check out how to solve log equations involving change of base

Click here for Contents Page

Product Law

$\ \log_a mn = \log_a m + \log_a n $

note that the logarithms are of the same base

Quotient Law

$\ \log_a {\frac{m}{n}} = \log_a m - \log_a n $

note that the logarithms are of the same base

Power Law

$\ \log_a m^n = n\log m $

Important: $\ \log_a xy^n \ne n\log_a xy $

The power n belongs to the base y only. The power n does not belong to base x

$ \begin{aligned}

\log_a xy^n &= \log_a x + \log_a y^n \quad \text{ applying power law} \\

\log_a xy^n &= \log_a x + n\log_a y \\

\end{aligned}$

Using the Laws of logarithms to solve equations

Strategy

- Obtain a single logarithm function on the LHS as well as the RHS
- Ensure that logarithm functions on the LHS and RHS have the same base
- Compare the arguments to solve for x

Example 1

$ \begin{aligned}

\lg {(5y +2)} &= 1 +\lg y \\

\lg {(5y +2)} &= \lg 10 +\lg y \qquad \text{Note that} 1 = \lg10 \\

\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\

5y + 2 &= 10y \qquad \text{(comparing arguments)} \\

5y &= 2 \\

y &= \frac{2}{5} \\

\end{aligned}$

Example 2

$ \begin{aligned}

\log_3 {(5y +7)} &= 2 +\log_3 {(y + 4)} \\

\log_3 {(5y +7)} &= 2\log_3{3} + \log_3 {(y + 4)} \qquad \text{Note that} 2 = 2\log_3{3} \\

\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\

5y + 2 &= 10y \qquad \text{(comparing arguments)} \\

5y &= 2 \\

y &= \frac{2}{5} \\

\end{aligned}$

Example 3

Solve the following equation

$\log(x+20) - \log(x+2) = \log x $

$\begin{aligned}

\log(x+20) - \log(x+2) &= \log x \\

\log \frac{ x+20}{ x+2 } &= \log x \\

\frac{ x+20}{ x+2 } &= x \\

x + 20 &= x^2 + 2x \\

x^2 + x - 20 &= 0 \\

(x + 5)( x - 4) &= 0 \\

x = -5 \; \text{rejected} \qquad x = 4 \\

\end{aligned} $

$ x = -5 $ is rejected as you cannot log a negative value

Example 4

Solve the following equation: $ 3^{x-4} = 5^{x-1} $

$\begin{aligned}

3^{x-4} &= 5^{x-1} \\

\log{3^{x-4}} &= \log{5^{x-1}} \qquad \text{log both sides} \\

(x-4) \log{3} &= (x-1) \log{5} \\

x( \log{3} - \log{5}) &= 4\log{3} - \log{5} \\

x &= \frac{4\log{3} - \log{5}}{ \log{3} - \log{5}} \\

\end{aligned} $

$ log_{10} 3 $ and $ log_{10} 5 $ can both be evaluated using the calculator

Take home message

To solve log equations:

- Obtain a single logarithm function on the LHS as well as the RHS
- Ensure that logarithm functions on the LHS and RHS have the same base
- Compare the arguments to solve for x

What's Next?

Click here to check out how to solve log equations involving change of base

Click here for Contents Page

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