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.