# How to Solve “Simple” Logarithmic Equations

We can solve basic logarithmic equations by equating the arguments (the expressions inside the parentheses) and then solving 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 \(b \ne 1\).

In some situations, we’ll need to convert a logarithmic equation into an exponential equation to find the solution. I’ll cover that in a future lesson.

**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.