Solving Exponential Equations with Logarithms

Hello! In our last lesson, we explored the properties of logarithms, learning how to expand and condense logarithmic expressions. You saw that the laws of logarithms are direct consequences of the laws of indices (exponents).

Today, we will use these properties to solve a new class of equations. The learning outcome for this lesson is to solve exponential equations using logarithms. These are equations where the variable you need to solve for, like $x$ , is in the exponent, such as in the equation $2^{x} = 11$ . We'll see how logarithms, and particularly the power rule, provide the key to "unlocking" the variable from its position in the exponent.

1. The Core Strategy: Isolating the Exponent

Because logarithms and exponentials are inverse operations, we can use logarithms to solve exponential equations. There are two slightly different but equivalent ways to think about this.

Let's take a simple equation: $5^{x} = 60$

Method 1: Convert to Logarithmic Form

The most direct method is to use the fundamental definition of a logarithm. Remember that the statement $a^{n} = b$ is equivalent to $lo g_{a} (b) = n$ .

Applying this to our problem:
$5^{x} = 60 ⟺ x = lo g_{5} (60)$

This gives us the exact answer. To find a decimal approximation, you can use a calculator. Many modern calculators have a button that looks like log□□ which allows you to input a custom base.

All of Logarithms in 25 Minutes!! | Chapter 14 (Part 1) | A Level Pure Maths

The following video from The GCSE Maths Tutor provides a clear, quick demonstration of this direct conversion method and how to use a calculator to find the answer.

Watch the segment from 07:03 to 08:51. It covers solving an equation similar to our example by converting it to logarithmic form.

Using a calculator, we find $x = lo g_{5} (60) \approx 2.544$ .

Method 2: Take the Logarithm of Both Sides

This method is more versatile and is essential for solving more complex equations. The strategy is to take a common logarithm of both sides of the equation. You can use either the common log (base 10, written as log) or the natural log (base e, written as ln). The result will be the same.

Let's solve $5^{x} = 60$ again using this method.

Take the log of both sides:
$lo g (5^{x}) = lo g (60)$
Apply the Power Rule: This is the crucial step. The power rule, $lo g_{a} (x^{k}) = k lo g_{a} (x)$ , allows us to bring the exponent down.
$x lo g (5) = lo g (60)$
Solve for x: Now $x$ is no longer in the exponent, and we can isolate it with simple division. $lo g (5)$ is just a number.
$x = \frac{l o g ( 60 )}{l o g ( 5 )}$

If you compute this on your calculator, you will get the same result: $x \approx \frac{1.778}{0.699} \approx 2.544$ .

This second method is powerful because it works for any exponential equation, no matter how complex the base or exponent.

Logarithms

For a clear text-based explanation and more examples of this method, please read the following short section from the mathcentre.ac.uk guide on logarithms.

Read 'Section 12. Using logarithms to solve equations' on page 8. It provides two excellent worked examples of this technique.

Test your understanding!

Solve the equation $7^{x} = 90$ for $x$ , giving your answer to three decimal places. Try solving it using both methods.

Show answer

Method 1 (Direct Conversion):
$7^{x} = 90 ⟹ x = lo g_{7} (90)$
Using a calculator, $x \approx 2.312$ .

Method 2 (Taking Logs):
\begin{align*} 7^x &= 90 \ \ln(7^x) &= \ln(90) \ x \ln(7) &= \ln(90) \ x &= \frac{\ln(90)}{\ln(7)} \approx \frac{4.4998}{1.9459} \approx 2.312 \end{align*}
Note: Using log (base 10) would yield the same final answer.

2. Equations with Variables in Both Exponents

What if the variable $x$ appears in exponents on both sides of the equation, with different bases? For example:
$7^{x + 1} = 3^{x + 2}$
Here, direct conversion isn't possible. The "take the log of both sides" method is the way to go.

Take the log of both sides:
$lo g (7^{x + 1}) = lo g (3^{x + 2})$
Apply the Power Rule: Remember to put brackets around the exponents.
$(x + 1) lo g (7) = (x + 2) lo g (3)$
Expand the brackets:
$x lo g (7) + lo g (7) = x lo g (3) + 2 lo g (3)$
Collect x terms on one side and constant terms on the other:
$x lo g (7) - x lo g (3) = 2 lo g (3) - lo g (7)$
Factorise (factor out) x:
$x (lo g (7) - lo g (3)) = 2 lo g (3) - lo g (7)$
Isolate x:
$x = \frac{2 l o g ( 3 ) - l o g ( 7 )}{l o g ( 7 ) - l o g ( 3 )}$

This is the exact answer. We can use the laws of logarithms from our last lesson to simplify it further before calculating:
$x = \frac{l o g ( 3 ^{2} ) - l o g ( 7 )}{l o g ( 7/3 )} = \frac{l o g ( 9 ) - l o g ( 7 )}{l o g ( 7/3 )} = \frac{l o g ( 9/7 )}{l o g ( 7/3 )}$
Using a calculator, $x \approx 0.339$ .

Edexcel AS Level Maths: 14.6 Solving Equations using Logarithms

This process involves several algebraic steps. Watch the following video for a very clear, guided example of solving this exact type of problem.

Watch the segment from 04:46 to 08:31. The presenter gives a step-by-step walkthrough, explaining each part of the algebraic manipulation.

3. Equations Reducible to Quadratic Form

Sometimes an exponential equation doesn't look straightforward, but it has a hidden structure. Consider the equation:
$3^{2 x} + 3^{x + 1} - 10 = 0$

Notice the relationship between the exponential terms. We can use the laws of indices to rewrite it:

$3^{2 x} = (3^{x})^{2}$
$3^{x + 1} = 3^{x} \cdot 3^{1} = 3 (3^{x})$

Substituting these back into the equation gives:
$(3^{x})^{2} + 3 (3^{x}) - 10 = 0$

This now looks like a quadratic equation. We can make this clearer with a substitution. Let $y = 3^{x}$ .

$y^{2} + 3 y - 10 = 0$

Now we can solve this quadratic for $y$ by factorising:
$(y + 5) (y - 2) = 0$
This gives two possible solutions for $y$ : $y = - 5$ or $y = 2$ .

Finally, we must substitute back to solve for $x$ .

Case 1: $y = - 5$
$3^{x} = - 5$ . This has no solution. A positive number raised to any real power can never be negative.
Case 2: $y = 2$
$3^{x} = 2$ . We can now solve this using our methods from earlier.
$x = lo g_{3} (2) \approx 0.631$

So, the only solution to the original equation is $x = lo g_{3} (2)$ .

Edexcel AS Level Maths: 14.6 Solving Equations using Logarithms

This technique of using a substitution to solve a quadratic-style exponential equation is very common. The same video resource provides two excellent worked examples.

Watch the first part of the video, from 00:11 to 04:46. It covers two examples, carefully explaining the substitution and the importance of checking for invalid solutions.

Test your understanding!

Solve the equation $e^{2 x} - 7 e^{x} + 12 = 0$ . (Hint: Let $y = e^{x}$ ).

Show answer

Substitute $y = e^{x}$ :
The equation becomes $y^{2} - 7 y + 12 = 0$ .
Solve the quadratic for y:
Factorising gives $(y - 3) (y - 4) = 0$ .
So, $y = 3$ or $y = 4$ .
Substitute back to solve for x:
- If $y = 3$ , then $e^{x} = 3$ . Taking the natural log of both sides gives $x = ln (3)$ .
- If $y = 4$ , then $e^{x} = 4$ . Taking the natural log of both sides gives $x = ln (4)$ .
Both solutions are valid since $e^{x}$ can equal 3 and 4.
The solutions are $x = ln (3) \approx 1.099$ and $x = ln (4) \approx 1.386$ .

Conclusion

In this lesson, we have seen how logarithms are the essential tool for solving exponential equations. By applying the laws of logarithms—especially the power rule—you can systematically solve for variables that are in an exponent.

Key Takeaways:

Simple Case ( $a^{x} = b$ ): You can either convert directly to logarithmic form ( $x = lo g_{a} b$ ) or take the log of both sides ( $x = \frac{l o g b}{l o g a}$ ).
Complex Case ( $a^{f (x)} = b^{g (x)}$ ): Always use the method of taking the logarithm of both sides. This allows you to use the power rule to bring the expressions $f (x)$ and $g (x)$ down, after which you can solve using algebra.
Quadratic Form: Look for equations that can be rewritten using a substitution (e.g., $y = a^{x}$ ) to form a quadratic equation.
Check for Validity: Always remember that $a^{x}$ must be positive for a positive base $a$ . This means you must discard any solutions that would require $a^{x}$ to be zero or negative.

Preview of the Next Lesson:

Having used logarithms to solve exponential equations, we will now flip the script. In the next lesson, we will solve logarithmic equations. This will often involve the inverse process: using the properties of logarithms to condense expressions and then converting the logarithmic equation into an exponential one to find the solution.

Recommended resources

1. The Core Strategy: Isolating the Exponent

2. Equations with Variables in Both Exponents

3. Equations Reducible to Quadratic Form

Conclusion