Explainer: Logarithmic Equations with Like Bases

In this explainer, we will learn how to solve logarithmic equations with like bases using the laws of exponents and logarithms.

Since the logarithm function to the base 𝑏 is inverse to the exponential function 𝑏𝑥, we will also use exponential functions to solve such equations. In essence, by the definition of log𝑏𝑥, it follows that given the number 𝐾 to solve log𝑏𝑥=𝐾, we must have 𝑏log𝑏𝑥=𝑏𝐾or𝑥=𝑏𝐾.

The first steps are usually to simplify the equations using the laws of logarithms. For example, so solve log6𝑥+log63=3 it is not clear what log63 is, so we recall that the sum of logarithms to a fixed base is just the log of a product, and rewrite the left as log6(3𝑥)=3 so that, exponentiating both sides to the 6th power, 3𝑥=63=216 and 𝑥=72.

Example 1: Solving Logarithmic Equations

Find 𝑥 such that log(4𝑥4)=2.


Since log means log10, the statement that log(4𝑥4)=2 is the same as 4𝑥4=102=1004𝑥=104𝑥=1044=26.

We must sometimes be wary of extraneous solutions.

Example 2: Solving Logarithmic Equations Where the Unknown Is in the Base of the Logarithm

What is the solution set of the equation log𝑥+264=2?


Here, the unknown is the base. Again, we rewrite this as an exponential equation 64=(𝑥+2)2, which becomes a quadratic: 𝑥2+4𝑥+4=64𝑥2+4𝑥60=0(𝑥+10)(𝑥6)=0 such that the solution set is apparently {6,10}.

But we are expecting 𝑥+2 to be positive since it is a logarithmic base. This is not true for 𝑥=10, which we dismiss as an extraneous solution. Therefore, the solution set is 6.

Example 3: Solving Logarithmic Equations

Find the solution set of log7𝑥2+log7𝑥2+1=log722 in .


This looks awkward because of the squared terms: there is no simplification of either log7𝑥2 or log722. So, we consider this all as an equation in 𝑦=log7𝑥, from which we get log7𝑥2+log7𝑥2+1=log722, so log7𝑥2+2log7𝑥+1=log722𝑦2+2𝑦+1=log722(𝑦+1)2=log722. This equation of squares is solved by 𝑦+1=log72or𝑦+1=log72. Rewriting these two equations using log7𝑥 once more, log7𝑥+1=log72 implies that log7𝑥log72=1log7𝑥2=1𝑥2=71=17𝑥=27 or log7𝑥+1=log72 implies that log7𝑥+log72=1log7(2𝑥)=12𝑥=71=17𝑥=114. The solution set is 27,114.

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