Welcome to Exponential and Logarithmic Equations

In your previous studies, you’ve learned how to solve equations where the variable is on the "ground floor," like \(2x = 8\). But what happens when the variable is hiding in the attic? Equations like \(2^x = 8\) are called exponential equations. In this chapter, we will learn how to use logarithms to bring those variables down to earth and solve for them. This is a vital skill for Further Pure Mathematics, as these equations appear in everything from population growth to radioactive decay!

1. Solving Equations of the Form \(a^x = b\)

This is the most fundamental type of exponential equation. There are two main ways to solve these, depending on whether the numbers are "friendly" with each other.

Method A: Using a Common Base

If you can write both sides of the equation using the same base, the solution is simple. You just compare the powers.

Example: Solve \(2^x = 32\).
Since we know that \(32 = 2^5\), we can write:
\(2^x = 2^5\)
Therefore, \(x = 5\).

Method B: Taking Logarithms of Both Sides

Most of the time, the numbers won't be so friendly (e.g., \(3^x = 20\)). When this happens, we use the Power Law of logarithms: \(\log_a(x^k) = k \log_a x\).

Step-by-Step Process:

  1. Take the logarithm of both sides (usually base 10, written as \(\lg\), or natural log \(\ln\)).
  2. Use the power law to move the variable \(x\) to the front.
  3. Divide to isolate \(x\).
  4. Use your calculator to find the final decimal value.

Example: Solve \(3^x = 20\).
\(\log(3^x) = \log(20)\)
\(x \log 3 = \log 20\)
\(x = \frac{\log 20}{\log 3}\)
\(x \approx 2.73\) (to 3 significant figures).

Quick Tip: It doesn't matter which base you use for the log as long as you use the same one on both sides. Most students find \(\log_{10}\) or \(\ln\) the easiest because they have dedicated buttons on the calculator!

2. Using the Change of Base Formula

Sometimes you will encounter equations with logarithms that have different bases. To solve these, you must make the bases match using the Change of Base Formula.

The formula from your syllabus is:
\(\log_a x = \frac{\log_b x}{\log_b a}\)

Another very useful version for "swapping" base and number is:
\(\log_a b = \frac{1}{\log_b a}\)

Example: Solve \(\log_2 x + \log_4 x = 6\).
We need a common base. Let's change \(\log_4 x\) into base 2:
\(\log_4 x = \frac{\log_2 x}{\log_2 4}\)
Since \(\log_2 4 = 2\), this becomes \(\frac{\log_2 x}{2}\) or \(\frac{1}{2} \log_2 x\).

Now the equation is:
\(\log_2 x + \frac{1}{2} \log_2 x = 6\)
\(\frac{3}{2} \log_2 x = 6\)
\(\log_2 x = 4\)
\(x = 2^4 = 16\).

Key Takeaway: Before adding or subtracting logs, always check if the bases are the same. If they aren't, "Change of Base" is your best friend!

3. "Hidden" Quadratic Equations

In Further Pure Maths, examiners love to disguise quadratic equations as exponential ones. These usually look like this: \(a^{2x} + b(a^x) + c = 0\).

How to spot them: Look for one term where the power is exactly double the power of another term (e.g., \(2^x\) and \(2^{2x}\)).

Step-by-Step Process:

  1. Use a substitution: Let \(y = a^x\).
  2. Rewrite the equation: This turns \(a^{2x}\) into \(y^2\).
  3. Solve the resulting quadratic equation for \(y\).
  4. Substitute back: Set \(a^x\) equal to your values of \(y\) and solve for \(x\).

Example: Solve \(2^{2x} - 5(2^x) + 4 = 0\).
Let \(y = 2^x\). The equation becomes:
\(y^2 - 5y + 4 = 0\)
\((y - 4)(y - 1) = 0\)
So, \(y = 4\) or \(y = 1\).

Now solve for \(x\):
If \(2^x = 4\), then \(x = 2\).
If \(2^x = 1\), then \(x = 0\).
Final answers: \(x = 2, x = 0\).

Common Mistake: Don't forget the last step! Students often find \(y\) and think they are finished. Always remember you are trying to find \(x\).

4. Simultaneous Equations

You may be asked to solve two equations with two unknowns (\(x\) and \(y\)) involving indices or logs. The goal is always to simplify them into linear equations first.

Strategy:
1. Use log laws or index laws to "remove" the logs/indices.
2. Solve the resulting simultaneous equations using substitution or elimination.

Example: Solve \(3^x \cdot 3^y = 27\) and \(\log_2 x - \log_2 y = 1\).
First equation: \(3^{x+y} = 3^3 \implies x + y = 3\)
Second equation: \(\log_2(\frac{x}{y}) = 1 \implies \frac{x}{y} = 2^1 \implies x = 2y\)

Substitute \(x = 2y\) into the first equation:
\(2y + y = 3\)
\(3y = 3 \implies y = 1\)
If \(y = 1\), then \(x = 2(1) = 2\).
Solution: \(x = 2, y = 1\).

Summary Checklist

  • Is the variable in the power? Use logs on both sides.
  • Are the log bases different? Use the Change of Base formula.
  • Does it look like a quadratic? Use the substitution \(y = a^x\).
  • Did you check your answers? Remember that you cannot take the log of a negative number! If you get a solution like \(2^x = -3\), there is no real solution for that part.

Don't worry if this seems tricky at first! The more you practice "bringing down the power," the more natural it will feel. Just remember your log laws from the previous chapter!

Quick Review of Log Laws needed for these equations:
\(\log_a(xy) = \log_a x + \log_a y\)
\(\log_a(\frac{x}{y}) = \log_a x - \log_a y\)
\(\log_a(x^k) = k \log_a x\)
\(\log_a a = 1\)
\(\log_a 1 = 0\)