More logarithmic equations

Exponential functions and logarithms: Logarithmic functions

More logarithmic equations

An important rule for solving logarithmic equations is the following rule.

\[\log_\blue{a}\left(\green{b}\right)=\log_\blue{a}\left(\purple{c}\right)\] gives \[\green{b}=\purple{c}\]

Example

\begin{array}{rcl}\log_\blue{2}\left(\green{x}\right)&=&3\\\log_\blue{2}\left(\green{x}\right)&=&\log_\blue{2}\left(\blue{2}^3\right)\\\green{x}&=&\blue{2}^3\\\green x &=&\purple{8}\end{array}

The example given is, of course, simplistic. It becomes interesting when we use this rule to solve equations of the following form.

Solve the equation for #x# :
\[
\log_{6}\left(x+5\right)=\log_{6}\left(25\right)
\]
Do not use exponents and give your answer in the form #x=\ldots#

#x=20#

\(\begin{array}{rcl}
\log_{6}\left(x+5\right)&=&\log_{6}\left(25\right)\\
&&\blue{\text{the original equation}}\\
x+5&=&25\\
&&\blue{\log_a\left(b\right)=\log_a\left(c\right)\text{ gives }b=c}\\
x&=&20\\
&&\blue{\text{moved constant term to the right}}
\end{array}\)

New example

With the help of calculation rules for logarithms we can rewrite more difficult equations as #\log_\blue{a}\left(\green{b}\right)=\log_\blue{a}\left(\purple{c}\right)#, which we can solve. In some cases, we will need to use the quadratic formula. If we do this, we must check whether the solutions are valid; the logarithm is not defined for negative numbers.

Solving logarithmic equations with calculation rules and the quadratic formula

	Step-by-step	Example
	We solve a logarithmic equation with unknown #x# with the help of the calculation rules for logarithms and the quadratic formula.	#2\cdot\log_2\left(x\right)=1+\log_2\left(x+4\right)#
Step 1	Write all terms as a logarithm.	#2\cdot \log_2\left(x\right)=\log_2\left(2\right)+\log_2\left(x+4\right)#
Step 2	Apply the rule #\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)=\log_{\blue{a}}\left(\green{b}^\purple{n}\right)#	#\log_2\left(x^2\right)=\log_2\left(2\right)+\log_2\left(x+4\right)#
Step 3	Take logarithms together so you get an equation of the form #\log_\blue{a}\left(\green{b}\right)=\log_\blue{a}\left(\purple{c}\right)#.	#\log_\blue{2}\left(\green{x^2}\right)=\log_\blue{a}\left(\purple{2x+8}\right)#
Step 4	Write the equation #\green{b}=\purple{c}#.	#\green{x^2}=\purple{2x+8}#
Step 5	Rewrite the equation #\green{b}=\purple{c}# as an equation you can solve using the quadratic formula.	#\green{x^2}-\purple{2x-8}=0#
Step 6	Solve the equation using the quadratic formula.	#x=4 \vee x=-2#
Step 7	Check if the solutions are valid	#x=4# gives #2\cdot\log_2\left(4\right)=1+\log_2\left(8\right)# so it is valid #x=-2# gives #2\cdot\red{\log_2\left(-2\right)}=1+\log_2\left(2\right)# so it is not valid

Solve the equation for #x# :
\[2\cdot \log_3\left(x\right)-4=\log_3\left(9 x+7\right)\]
Write:

#none# if there is no solution,
#x=x_1# if there is one solution,
#x=x_1\lor x=x_2# if there are two solutions,

with the correct exact values #x_1# and #x_2#.

#x=\frac{9+\sqrt{{{6589}\over{81}}}}{{{2}\over{81}}}#

We use the step-by-step guide.

Step 1	Write all terms as a logarithm. \[\begin{array}{rcl}2\cdot \log_3\left(x\right)-4&=&\log_3\left(9 x+7\right)\\&&\blue{\text{the original equation}}\\2\cdot \log_3\left(x\right)-\log_{3}\left(3^{4}\right)&=&\log_3\left(9 x+7\right)\\&&\blue{b=\log_a\left(a^b\right)}\\2\cdot \log_3\left(x\right)-\log_{3}\left(81\right)&=&\log_3\left(9 x+7\right)\\&&\blue{\text{calculated exponents}}\end{array}\]
Step 2	Apply the rule #\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)=\log_{\blue{a}}\left(\green{b}^\purple{n}\right)#. \[\begin{array}{rcl}\log_3\left(x^2\right)-\log_{3}\left(81\right)&=&\log_3\left(9 x+7\right)\\&&\blue{n\cdot \log_a\left(b\right)=\log_a\left(b^n\right)}\end{array}\]
Step 3	Take logarithms together so you get an equation of the form #\log_\blue{a}\left(\green{b}\right)=\log_\blue{a}\left(\purple{c}\right)#. \[\begin{array}{rcl}\log_\blue{3}\left(\green{\frac{1}{81} x^2}\right)&=&\log_\blue{3}\left(\purple{9 x+7}\right)\\&&\blue{\log_a\left(b\right)-\log_a\left(c\right)=\log_a\left(\frac{b}{c}\right)}\end{array}\]
Step 4	Write the equation #\green{b}=\purple{c}#. \[\begin{array}{rcl}\green{\frac{1}{81} x^2}&=&\purple{9 x+7}\end{array}\]
Step 5	Rewrite the equation #\green{b}=\purple{c}# as an equation you can solve using the quadratic formula. \[\begin{array}{rcl}\green{\frac{1}{81} x^2}-\purple{9 x-7}&=&0\end{array}\]
Step 6	Solve the equation using the quadratic formula. \[x_1=\frac{9+\sqrt{{{6589}\over{81}}}}{{{2}\over{81}}}\vee x_2=\frac{9-\sqrt{{{6589}\over{81}}}}{{{2}\over{81}}}\]
Step 7	Check if the solutions are valid. #x_1# is positive, so if we substitute #x_1# we'll only get positive values. Therefore #x_1# is valid! We see that #x_2# is a negative value. If we were to substitute this, we would get a negative logarithm on the left hand side of the equation, which is not allowed. Therefore, only #x_1# is valid.

New example