Solving Exponential and Logarithmic Equations

An exponential equation hides the unknown in an exponent, like \(b^x = k\); a logarithmic equation hides it inside a log, like \(\log_b(x) = k\). Both are solved the same way: apply the inverse operation to both sides.

By the end you'll be able to solve \(b^x = k\) by taking a log of both sides, solve \(\log_b(x) = k\) by exponentiating both sides, and catch extraneous solutions that don't belong.

Predict: in log form, what is the solution of \(2^x = 10\)? Write down your guess, then press Animate below to watch the log get applied and check it against the graph.

The curve is \(y=b^x\) and the dashed line is \(y=k\) — wherever they cross IS the solution. Taking log of both sides turns the exponent into a coefficient (the power rule), so you can isolate x algebraically instead of guessing. Switch to the log-equation mode to see the mirror-image move: exponentiating both sides undoes a log.

\(x = \log(10)/\log(2) \approx\) 3.32

y = b^x and y = k — they cross at x
y = b^x y = k (target) solution

Exponentials and logs are inverses, so every exponential or logarithmic equation is solved by applying the OTHER operation to both sides, then cleaning up with the log properties.

Procedural

Two strategies, depending on whether the bases match:

  • Same base: if both sides can be written with the same base, set the exponents equal. \(2^{x+1} = 8 = 2^3\) gives \(x+1=3\), so \(x=2\) — no log needed.
  • Take a log: when the bases can't be matched, take log or ln of both sides and use the power rule \(\log(b^x)=x\log b\) to bring the exponent down as a coefficient, then divide to isolate x. The change-of-base formula is exactly this last division, letting any base be evaluated on a calculator.
  • Logarithmic equations go the other way: rewrite \(\log_b(x)=k\) in exponential form \(x = b^k\), or condense multiple logs into one first using the product/quotient rules.
Formal

The power rule \(\log(b^x) = x\log b\) — the same rule from logarithm properties — is what makes step 2 of the graph above legal: it's the one move that pulls x out of the exponent and into an ordinary coefficient, turning an exponential equation into a linear one. Without it, \(b^x=k\) would have no clean algebraic path to x at all.

Whenever you solve a logarithmic equation, always check that every logarithm's argument is positive in the ORIGINAL equation — algebra can introduce an extraneous solution that solves the transformed equation but not the real one.

Applied

"How many years until this investment doubles?" and "how many bacteria generations until the population hits a million?" are both exponential equations in disguise — you isolate the unknown time or count by taking a log, exactly like the slider demo above.

Worked example

Solve \(3^x = 20\). Take the natural log of both sides: \(\ln(3^x) = \ln(20)\). Power rule: \(x\ln 3 = \ln 20\). Divide: \(x = \ln(20)/\ln(3) \approx 2.9957/1.0986 \approx\) 2.727.

A logarithmic example: solve \(\log_2(x-1) = 4\). Rewrite in exponential form: \(2^4 = x-1\), so \(16 = x-1\) and \(x=17\). Check: \(x-1 = 16 > 0\), so the argument is positive and the answer is valid.

Your turn

Solve \(4^x = 35\). Take the log of both sides: \(\log(4^x) = \log(35)\). Power rule: \(x\log 4 = \log 35\). Divide: \(x = \log(35)/\log(4) \approx 1.5441/0.6021 = \) ____

Reveal the answer

\(x \approx\) 2.565. Set b = 4 and k = 35 on the slider above (mode "Solve \(b^x=k\)") and check that the readout matches — the curve crosses y = 35 at x ≈ 2.565.

More info — why taking a log of both sides is legal

\(\log(x)\) is a strictly increasing function, so it's one-to-one: if \(A = B\), then \(\log(A) = \log(B)\) is guaranteed, and no solutions get created or lost by the step. That's exactly what licenses "take log of both sides" as a valid algebraic move, the same way squaring both sides is only safe when you check for sign issues afterward. See the Paul's Online Notes link in Dive deeper for more worked exponential and logarithmic equations.

Check your understanding

Question 1 of 4

Solve \(7^x = 50\) for x (round to three decimals). Which of these is the correct approach and answer?

Question 2 of 4

Which of these equations can be solved by matching bases — no logarithm needed at all?

Question 3 of 4

To solve \(6^x = 45\) you take the log of both sides to get \(\log(6^x) = \log(45)\). Which log property lets you rewrite the left side as \(x \cdot \log 6\), so you can isolate x?

Question 4 of 4

Solving \(\log_2(x) + \log_2(x-2) = 3\) with the product rule gives \(\log_2[x(x-2)] = 3\), so \(x(x-2)=8\), \(x^2-2x-8=0\), \((x-4)(x+2)=0\), so \(x=4\) or \(x=-2\). Which value(s) actually solve the ORIGINAL equation?

Recap

  • Same base on both sides? Set the exponents equal — no log needed.
  • Different bases: take log (or ln) of both sides, then use the power rule \(\log(b^x)=x\log b\) to bring the exponent down: \(x = \log(k)/\log(b)\).
  • \(\log_b(x)=k\) is solved by rewriting in exponential form: \(x = b^k\).
  • Always check candidate solutions against the original equation — a log's argument must be positive, or the solution is extraneous.

Dive deeper

Sources

  • Solving Exponential and Logarithmic Equations