We can solve simple logarithmic equations by setting the arguments (stuff inside the parenthesis) equal to each other, then solve for the variable.
We can do that because of the one-to-one property of logarithms as shown below.
If $${\log _b}\left( M \right) = {\log _b}\left( N \right)$$ then $$M = N$$
Assume that \(M\), \(N\), and \(b\) are positive real numbers but \(b \ne 1\).
There are cases where we will have to rewrite a logarithmic equation into an exponential equation to solve it. There is a separate lesson for that which can be found here.
Example 1: Solve the logarithmic equation below.
$${\log _2}\left( {3x + 16} \right) = {\log _2}\left( {5 \,-\, {{2x} \over 3}} \right)$$
STEP 1: Set each expression inside the parenthesis equal to each other.
$$3x + 16 = 5\, -\, {{2x} \over 3}$$
STEP 2: Move the variables to the left and the constants to the right.
$$3x + {{2x} \over 3} = 5 - 16$$
STEP 3: Solve for the variable \(x\).
$$\eqalign{
{{9x + 2x} \over 3} &= - 11 \cr
11x &= - 33 \cr
x &= - 3 \cr} $$
The final answer is \(\boxed{x = - 3}\). I encourage you to check the value of x with the original logarithmic equation.
Example 2: Solve the logarithmic equation below.
$$\log \left( {2x - 5} \right) + \log \left( {x + 4} \right) = \log \left( {3x - 2} \right)$$
STEP 1: Combine the logarithms on the left side using the the Product Property of Logarithms. It states that the log of a product is the sum of the logs.
$$\log \left[ {\left( {2x - 5} \right)\left( {x + 4} \right)} \right] = \log \left( {3x - 2} \right)$$
STEP 2: Multiply the binomials on the left side.
$$\log \left( {2{x^2} + 3x - 20} \right) = \log \left( {3x - 2} \right)$$
STEP 3: Set the arguments equal to each other then solve.
$$\eqalign{
2{x^2} + 3x - 20 &= 3x - 2 \cr
2{x^2} + 3x - 3x - 20 + 2 &= 0 \cr
2{x^2} - 18 &= 0 \cr
{x^2} - 9 &= 0 \cr} $$
STEP 4: Factor the binomial then set each factor equal to zero to solve for \(x\).
$$\eqalign{
\left( {x + 3} \right)\left( {x - 3} \right) &= 0 \cr
{x_1} &= - 3 \cr
{x_2} &= 3 \cr} $$
STEP 5: Check the two possible answers with the original equation to get rid of extraneous solutions.
For \({x_1} = - 3\):
$$\eqalign{
\log \left[ {2\left( { - 3} \right) - 5} \right] + \log \left[ {\left( { - 3} \right) + 4} \right] &= \log \left[ {3\left( { - 3} \right) - 2} \right] \cr
\log \left( { - 11} \right) + \log \left( 1 \right) &= \log \left( { - 11} \right) \cr} $$
Since we arrived at a case where we are taking the logarithm of a negative number, therefore \({x_1} = - 3\) is not a solution.
For \({x_2}=3\):
$$\eqalign{
\log \left[ {2\left( 3 \right) - 5} \right] + \log \left[ {\left( 3 \right) + 4} \right] &= \log \left[ {3\left( 3 \right) - 2} \right] \cr
\log \left( 1 \right) + \log \left( 7 \right) &= \log \left( 7 \right) \cr
\log \left( 7 \right) &= \log \left( 7 \right) \cr} $$
Since we arrived at a true statement, that means \({x_2}=3\) is a solution to the logarithmic equation.
Therefore, our solution is just \(\boxed{x=3}\).
Example 3: Solve the logarithmic equation below.
$$\ln \left( {2x - 1} \right) - \ln x = \ln \left( {{1 \over {x + 1}}} \right)$$
STEP 1: Combine the logarithms on the left side using the Quotient Property of Logarithms. It states that the quotient of logs is the difference of the logs.
$$\ln \left( {{{2x - 1} \over x}} \right) = \ln \left( {{1 \over {x + 1}}} \right)$$
STEP 2: Equate the arguments then solve for \(x\).
$$\eqalign{
{{2x - 1} \over x} &= {1 \over {x + 1}} \cr
2{x^2} + x - 1 &= x \cr
2{x^2} - 1 &= 0 \cr
2{x^2} &= 1 \cr
{x^2} &= {1 \over 2} \cr
x &= {{ \pm \sqrt 2 } \over 2} \cr
{x_1} &= {{\sqrt 2 } \over 2} \cr
{x_2} &= {{ - \sqrt 2 } \over 2} \cr} $$
STEP 3: You should verify that there is only one solution which is \(\boxed{x = {{\sqrt 2 } \over 2}}\). The negative value of \(x\) is an extraneous solution so disregard it.