How to Solve Exponential Equations?


By Robert O

In this lesson, you will learn straightforward ways to solve exponential equations. It is easy to convert exponential equations to logarithmic equations. After conversion, you apply the laws of logs to solve the resulting equation. In other words, solving exponential equations involves conversion to logarithms as the first step.

With exponential laws alone, you cannot solve some of the complex exponential problems. But with logarithms, everything turns out to be simple. Let’s remind ourselves about the log laws before looking at the examples.

What Are Log Laws And Their Properties?

The following are the most common laws and properties of logarithms:

    \[log _{b} 1=0, \text { from the exponential rule of } b^{0}=1\]

    \[log _{b} b=1, \text { from the exponential rule of } b^{1}=b\]

    \[log _{b} b^{x}=x . \text { In general } log _{b} b^{f(x)}=f(x)\]

    \[b^{log _{b} x}=x . \text { In general } b^{log _{b} f(x)}=f(x)\]

    \[\text { If } log _{b} x=log _{b} y, \text { then } x=y\]

Product rule:

    \[log _{b}(m . n)=log _{b}(m)+log _{b}(n)\]

Quotient rule:

    \[\log _{b}\left(\frac{m}{n}\right)=\log _{b}(m)-\log _{b}(n)\]

Power rule:

    \[log _{b}\left(m^{p}\right)=p log _{b}(m)\]

Change of base rule:

    \[log _{b} x=\frac{log _{a} x}{log _{a} b}\]

We now have the confidence to face any problems involving exponential equations because we have all it takes to find solutions.

Example 1

What is the value of x in the equation

    \[2^{x-1}=3^{x+1} ?\]


Start by introducing log, taking one of the bases that you have.

    \[log _{2} 2^{x-1}=log _{2} 3^{x+1}\]

Using power rule:

    \[(x-1) log _{2} 2=(x+1) log _{2} 3\]

    \[(x-1)=(x+1) log _{2} 3, \text { since } log _{2} 2=1\]

Applying change of base rule to the right side of the equation:

Change of base rule:

    \[log _{b} x=\frac{log _{a} x}{log _{a} b}\]

    \[x-1=(x+1) \frac{log _{10} 3}{log _{10} 2}\]

Evaluate logs and solve for x:

    \[x-1=(x+1) \frac{0.4771}{0.3010}\]

    \[x-1=(x+1) 1.585\]

    \[x-1=1.585 x+1.585\]


By substituting the value of x in the original equation, we conclude that we have a valid value.

Example 2

What is the value of x that satisfies the equation

    \[2^{1-x}=3^{x-1} ?\]


We start by introducing the logarithms to base 2 or base 2. In this case, we will take the logarithm to base 2.

    \[log _{2} 2^{1-x}=log _{2} 3^{x-1}\]

By applying the power rule:

    \[(1-x) log _{2} 2=(x-1) log _{2} 3\]

    \[(1-x)=(x-1) log _{2} 3\]

Using change of base rule:

    \[(1-x)=(x-1) \frac{log _{10} 3}{log _{10} 2}\]

By evaluating the logs and solving for x:

    \[(1-x)=(x-1) \frac{0.4771}{0.3010}\]

    \[(1-x)=(x-1) 1.585\]

    \[(1-x)=1.585 x-1.585\]

    \[-2.585 x=-2.585\]



The steps that you need to solve exponential equations are only three. Introduce logarithms to one of the bases, apply log laws to simplify the equation, and solve for x. That means you should have the laws of logarithms at the tips of your fingers if you need to save time and solve exponential equations correctly.

About the Author

This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.