Solving Exponential Equations & Inequalities

Q: How do I know which method to use?

Look at the numbers. If both sides can be written as powers of one base ($3^x = 81$, or $4^x = 8$ via base $2$), match the bases. If not ($5^x = 40$), take a log of both sides. When in doubt, the log method always works — matching bases is just a shortcut for nice numbers.

Q: Does it matter which log base I use in Method 2?

No. Any base gives the same answer: $\dfrac{\log 40}{\log 5} = \dfrac{\ln 40}{\ln 5}$. Use $\log$ or $\ln$ since both are on your calculator. If you use the base of the equation itself, the answer is even cleaner: $x = \log_5 40$.

Q: What if the exponential isn't alone, like $3 \cdot 2^x = 48$?

Isolate the exponential first, before either method. Divide both sides by $3$ to get $2^x = 16$, then proceed: $16 = 2^4$, so $x = 4$. Taking a log before isolating is the classic mistake — $\log(3 \cdot 2^x)$ is not $3 \cdot \log 2^x$.

An exponential equation has the variable in the exponent — things like $3^x = 81$ or $2^{x-1} = 16$ . That placement is what makes them feel different: you can't isolate $x$ by adding or dividing, because $x$ isn't a term or a factor. It's a power.

There are exactly two methods, and which one you use depends on the numbers. If both sides can be written as powers of the same base, match the bases and set the exponents equal — fast and clean. If they can't, take a logarithm of both sides and use the power rule to bring $x$ down out of the exponent. Every exponential equation you'll see falls into one of these two buckets.

Method 1: match the bases

If you can rewrite both sides as powers of one common base, the equation $b^{\text{left}} = b^{\text{right}}$ forces the exponents to be equal — an exponential function never repeats an output, so equal outputs mean equal inputs. From there it's ordinary algebra.

For $3^x = 81$ : recognize $81 = 3^4$ , so $3^x = 3^4$ and $x = 4$ . The skill this method really tests is knowing your powers — $2^5 = 32$ , $3^4 = 81$ , $5^3 = 125$ , and so on. When a problem's numbers are "nice," this is almost always the intended route.

Watch for hidden common bases. In $4^x = 8$ , neither number is a power of the other, but both are powers of $2$ : $4 = 2^2$ and $8 = 2^3$ . Rewriting gives $2^{2x} = 2^3$ , so $2x = 3$ and $x = \dfrac{3}{2}$ .

Method 2: take a logarithm of both sides

When the two sides don't share a base — like $5^x = 40$ , since $40$ is not a whole-number power of $5$ — take a log of both sides. Any base works; $\log$ (base $10$ ) or $\ln$ are the ones on your calculator.

The whole point is the power rule: $\log 5^x = x \log 5$ . Taking the log moves $x$ from the exponent down to a coefficient, and once $x$ is a coefficient, you divide: $x = \dfrac{\log 40}{\log 5}$ . The logarithm is the inverse of the exponential, which is exactly why it can pull the variable out of the exponent.

The exact answer is the log expression itself. Only round at the very end: $\dfrac{\log 40}{\log 5} \approx 2.29$ . And note the expression is $\log 40$ divided by $\log 5$ — not $\log \dfrac{40}{5}$ . Those are different numbers.

Choosing a method, and one warning

Scan the numbers first. Powers of a common base on both sides? Method 1. Anything else — Method 2. Both methods are correct on every solvable equation; matching bases is just quicker when it's available.

One warning for inequalities: if the base is between $0$ and $1$ , the function decreases, so comparing exponents flips the inequality sign. $\displaystyle \left(\frac{1}{2}\right)^x < \left(\frac{1}{2}\right)^4$ means $x > 4$ , not $x < 4$ . For plain equations there's nothing to flip — equal is equal.

Worked examples

Example 1: same base, simple exponent

Solve $3^x = 81$ .

Write

81

as a power of

3

81 = 3^4

Now both sides share base

3

3^x = 3^4

Set the exponents equal

x = 4

Check:

3^4 = 81

✓

Answer: $x = 4$

Example 2: same base, expression in the exponent

Solve $2^{x-1} = 16$ .

Write

16

as a power of

2

16 = 2^4

Match the bases

2^{x-1} = 2^4

Set the exponents equal

x - 1 = 4

Add

1

to both sides

x = 5

Answer: $x = 5$

Example 3: a hidden common base

Solve $4^x = 8$ .

Write both sides as powers of

2

4 = 2^2, \quad 8 = 2^3

Rewrite the equation

(2^2)^x = 2^3

Power to a power: multiply exponents

2^{2x} = 2^3

Set the exponents equal and solve

2x = 3, \quad x = \dfrac{3}{2}

Answer: $x = \dfrac{3}{2}$

Example 4: no common base — use logs

A bacteria culture multiplies by $5$ every day, so its size relative to the start is $5^x$ after $x$ days. Solve $5^x = 40$ . Round to the nearest hundredth.

40

is not a whole-number power of

5

, so take the log of both sides

\log 5^x = \log 40

Power rule: bring the exponent down

x \log 5 = \log 40

Divide both sides by

\log 5

x = \dfrac{\log 40}{\log 5}

Evaluate and round

x \approx 2.29

Answer: $x = \dfrac{\log 40}{\log 5} \approx 2.29$ days

Try one yourself

Common questions

How do I know which method to use?

Look at the numbers. If both sides can be written as powers of one base ( $3^x = 81$ , or $4^x = 8$ via base $2$ ), match the bases. If not ( $5^x = 40$ ), take a log of both sides. When in doubt, the log method always works — matching bases is just a shortcut for nice numbers.

Does it matter which log base I use in Method 2?

No. Any base gives the same answer: $\dfrac{\log 40}{\log 5} = \dfrac{\ln 40}{\ln 5}$ . Use $\log$ or $\ln$ since both are on your calculator. If you use the base of the equation itself, the answer is even cleaner: $x = \log_5 40$ .

What if the exponential isn't alone, like

3 \cdot 2^x = 48

Isolate the exponential first, before either method. Divide both sides by $3$ to get $2^x = 16$ , then proceed: $16 = 2^4$ , so $x = 4$ . Taking a log before isolating is the classic mistake — $\log(3 \cdot 2^x)$ is not $3 \cdot \log 2^x$ .

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