Solving Exponential Equations with Logarithms

An exponential equation has its variable sitting in an exponent, such as \(2^x = 10\) or \(3 \cdot 5^{x+1} = 60\). When the bases on each side of an equation cannot be matched (for example, \(2^x = 10\) has no simple power of 2 equal to 10), you cannot solve it by comparing exponents. Instead, you take the logarithm of both sides. Because logarithms are the inverse operation of exponentiation, this lets you pull the variable out of the exponent and solve for it algebraically.

The key log rule used here is the power rule: \(\log(a^x) = x\log(a)\). This rule lets an exponent become a multiplier once you apply a logarithm. If you already know how to convert a logarithm to exponential form, this is really the same relationship in reverse: instead of rewriting \(\log_b(y) = x\) as \(b^x = y\), you start with \(b^x = y\) and apply a log to both sides to recover \(x\).

You can use any base of logarithm you like, since it will cancel out on both sides of the equation, but \(\log_{10}\) (written \(\log\)) or the natural logarithm \(\ln\) (base \(e\)) are the most convenient because calculators compute them directly.

Follow these steps whenever the bases in an exponential equation cannot be made equal:

1. Isolate the exponential expression on one side of the equation.
2. Take \(\log\) or \(\ln\) of both sides.
3. Use the power rule to bring the exponent down as a multiplier: \(\log(b^x) = x\log(b)\).
4. Solve the resulting linear equation for the variable.
5. Use a calculator to find a decimal approximation if needed, and check the answer in the original equation.

Solve \(2^x = 10\).

Since 10 is not a power of 2, take the logarithm of both sides:

\(\log(2^x) = \log(10)\)

Apply the power rule:

\(x\log(2) = \log(10)\)

Divide both sides by \(\log(2)\):

\(x = \dfrac{\log(10)}{\log(2)} \approx 3.32\)

The graph below shows \(y = 2^x\) crossing \(y = 10\) at this same \(x\)-value, which is the solution to the equation.

Solve \(5 \cdot 3^{x} - 4 = 41\).

First isolate the exponential part:

\(5 \cdot 3^{x} = 45\)

\(3^{x} = 9\)

Here the bases actually do match, since \(9 = 3^2\), so \(x = 2\) without needing a logarithm at all. This example is a reminder to always isolate the exponential expression first: sometimes doing so reveals a matching base, and a logarithm is not even necessary.

Solve \(e^{2x+1} = 15\).

Take the natural logarithm of both sides, since the base is \(e\):

\(\ln(e^{2x+1}) = \ln(15)\)

Because \(\ln(e^{k}) = k\), the left side simplifies directly:

\(2x + 1 = \ln(15)\)

Solve for \(x\):

\(x = \dfrac{\ln(15) - 1}{2} \approx 0.85\)

Substituting this value back into \(e^{2x+1}\) confirms it produces approximately 15, so the solution checks out.

• Taking a logarithm before isolating the exponential term, which mixes up terms that are not actually being multiplied.
• Forgetting to apply the power rule correctly: \(\log(b^x) = x\log(b)\), not \(x \cdot \log(b^x)\) or similar mix-ups.
• Trying to take a logarithm of a negative number or zero. If isolating the exponential term leaves a non-positive value on the other side, the equation has no real solution.
• Rounding too early. Keep exact logarithmic expressions like \(\dfrac{\log(10)}{\log(2)}\) until the final step, then round.

Solving an exponential equation with logarithms is closely related to, but not the same as, solving logarithmic equations, where the variable sits inside a logarithm instead of an exponent. Both rely on the same inverse relationship between exponentials and logarithms, so building a solid understanding of what a logarithm is makes both types of equations much easier to work through.

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