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Exponential and logarithmic equations

Algebra · level 8, lesson 5 of 8

A question to start

The Perseverance rover on Mars keeps warm and makes its electricity from the heat of plutonium-238, which has a half-life of about 87.7 years: every 87.7 years, half of what is left decays. When will only three quarters of the plutonium be left?

The amount left after \(t\) years is a fraction \(\left(\tfrac12\right)^{t/87.7}\) of the start, so we need

\[\left(\tfrac12\right)^{t/87.7} = 0.75.\]

The unknown is in the exponent, and 0.75 is not a neat power of \(\tfrac12\), so the common-base trick from level 7 fails. Logarithms solve it in one line. (The answer is about 36.4 years.)

The idea

Method 1: common base (when it works)

If both sides can be written as powers of the same base, set the exponents equal. This works because \(b^x\) is one-to-one. \(9^{x+1} = 27^x\) becomes \(3^{2x + 2} = 3^{3x}\), so \(x = 2\).

Method 2: take logarithms

When there is no common base, take the logarithm of both sides. The power law \(\log\left(b^x\right) = x\log b\) brings the unknown down from the exponent:

\[3^x = 20 \;\Rightarrow\; \log\left(3^x\right) = \log 20 \;\Rightarrow\; x\log 3 = \log 20 \;\Rightarrow\; x = \frac{\log 20}{\log 3} \approx 2.727.\]

Why is it allowed to "take log of both sides"? If two positive numbers are equal, their logarithms are equal, because \(\log\) is a function (one input, one output). Always isolate the power first: in \(5 \cdot 3^x = 100\), divide by 5 before taking logs.

Method 3: a quadratic in disguise

\(4^x = \left(2^2\right)^x = \left(2^x\right)^2\). So in \(4^x - 3 \cdot 2^x - 4 = 0\), put \(u = 2^x\):

\[u^2 - 3u - 4 = 0 \;\Rightarrow\; (u - 4)(u + 1) = 0 \;\Rightarrow\; u = 4 \text{ or } u = -1.\]

\(2^x = 4\) gives \(x = 2\). \(2^x = -1\) is impossible, since \(2^x > 0\). So \(x = 2\) only.

Logarithmic equations

To solve an equation with logs of the unknown:

  1. Use the laws to combine into a single log on one side.
  2. Rewrite in exponential form: \(\log_b(\text{stuff}) = c\) means \(\text{stuff} = b^c\).
  3. Solve.
  4. Check every answer in the original equation. Each log needs a positive input.

Why extraneous roots appear

The law \(\log x + \log(x - 3) = \log\left(x(x - 3)\right)\) is only true when \(x > 0\) and \(x - 3 > 0\). But \(\log\left(x(x - 3)\right)\) also makes sense when both \(x\) and \(x - 3\) are negative (their product is positive). So combining the logs quietly widens the domain, and the new equation can have solutions the old one does not. Those are extraneous roots, and checking catches them.

Modelling with logs

Logarithmic scales measure things that range over many powers of ten.

  • pH (acidity): \(\text{pH} = -\log[\text{H}^+]\), where \([\text{H}^+]\) is the concentration of hydrogen ions in moles per litre. Pure water has pH 7; lower is more acidic. Each step of 1 is a factor of 10.
  • Decibels (loudness): \(L = 10\log\dfrac{I}{I_0}\), where \(I\) is the sound intensity and \(I_0\) is the quietest sound a person can hear. Each extra 10 dB multiplies the intensity by 10.
  • Half-life and doubling time: \(N = N_0\left(\tfrac12\right)^{t/H}\) and \(N = N_0 \cdot 2^{t/D}\). Logs find \(t\), \(H\) or \(D\).
years percent left 100 200 300 25 50 75 100 36.4 years: 75% 87.7 years: 50% 175.4 years: 25%

Worked examples

1. Taking logs. Solve \(3^x = 20\).

  • \(x = \dfrac{\log 20}{\log 3} \approx \dfrac{1.30103}{0.47712} \approx 2.727\). Check: \(3^{2.727} \approx 20\), and it lies between \(3^2 = 9\) and \(3^3 = 27\). ✓

2. Unknown on both sides. Solve \(2^{x + 1} = 3^x\).

  • Take logs: \((x + 1)\log 2 = x\log 3\).
  • Expand and collect the \(x\) terms: \(x\log 2 + \log 2 = x\log 3\), so \(\log 2 = x(\log 3 - \log 2)\).
  • \(x = \dfrac{\log 2}{\log 3 - \log 2} = \dfrac{\log 2}{\log 1.5} \approx 1.710\).

3. A log equation with an extraneous root. Solve \(\log x + \log(x - 3) = 1\).

  • Combine: \(\log\left(x(x - 3)\right) = 1\). Exponential form: \(x(x - 3) = 10^1 = 10\).
  • \(x^2 - 3x - 10 = 0\), so \((x - 5)(x + 2) = 0\), so \(x = 5\) or \(x = -2\).
  • Check \(x = 5\): \(\log 5 + \log 2 = \log 10 = 1\). ✓ Check \(x = -2\): \(\log(-2)\) does not exist. Reject.
  • Answer: \(x = 5\).

4. Logs on both sides. Solve \(\log_2(x + 1) - \log_2(x - 1) = 3\).

  • \(\log_2\dfrac{x + 1}{x - 1} = 3\), so \(\dfrac{x + 1}{x - 1} = 8\).
  • \(x + 1 = 8x - 8\), so \(7x = 9\), so \(x = \dfrac97\).
  • Check: \(x - 1 = \tfrac27 > 0\) and \(x + 1 = \tfrac{16}{7} > 0\). ✓

5. The rover. Solve \(\left(\tfrac12\right)^{t/87.7} = 0.75\).

  • Take logs: \(\dfrac{t}{87.7}\log 0.5 = \log 0.75\).
  • \(t = 87.7 \times \dfrac{\log 0.75}{\log 0.5} \approx 87.7 \times 0.4150 \approx 36.4\) years.

6. Carbon dating. Carbon-14 has a half-life of about 5730 years. A bone has 30 percent of the carbon-14 it had when the animal died. How old is it?

  • \(\left(\tfrac12\right)^{t/5730} = 0.3\), so \(t = 5730 \times \dfrac{\log 0.3}{\log 0.5} \approx 5730 \times 1.737 \approx 9950\) years.

7. pH and decibels. Lemon juice has a pH of about 2, and pure water 7. How many times more hydrogen ions does lemon juice have?

  • \([\text{H}^+] = 10^{-\text{pH}}\), so the ratio is \(\dfrac{10^{-2}}{10^{-7}} = 10^5 = 100\,000\) times.
  • Similarly, a 90 dB sound is \(10^{(90 - 60)/10} = 10^3 = 1000\) times more intense than a 60 dB sound.

Try it

  1. Solve \(2^x = 50\) to 4 decimal places.
  2. Solve \(6^{2x - 1} = 30\).
  3. Solve \(\log_3(x + 4) = 2\).
  4. Solve \(\log_2 x + \log_2(x + 2) = 3\).
  5. Solve \(9^x - 4 \cdot 3^x + 3 = 0\).
  6. A culture grows from 1000 to 5000 bacteria in 6 hours. What is its doubling time?
  7. How many times more intense is a 100 dB sound than a 70 dB sound?
  8. Solve \(2\ln x = \ln(3x + 4)\).
Answers
  1. \(x = \dfrac{\log 50}{\log 2} \approx 5.6439\).
  2. \((2x - 1)\log 6 = \log 30\), so \(2x - 1 = \dfrac{\log 30}{\log 6} \approx 1.8982\), so \(x \approx 1.4491\).
  3. \(x + 4 = 3^2 = 9\), so \(x = 5\). Check: \(\log_3 9 = 2\). ✓
  4. \(\log_2\left(x(x + 2)\right) = 3\), so \(x^2 + 2x = 8\), so \((x + 4)(x - 2) = 0\). \(x = -4\) makes \(\log_2 x\) undefined: reject. \(x = 2\): \(\log_2 2 + \log_2 4 = 1 + 2 = 3\). ✓ Answer \(x = 2\).
  5. Put \(u = 3^x\): \(u^2 - 4u + 3 = 0\), so \(u = 1\) or \(u = 3\). \(3^x = 1\) gives \(x = 0\); \(3^x = 3\) gives \(x = 1\).
  6. \(5000 = 1000 \cdot 2^{6/D}\), so \(2^{6/D} = 5\), so \(\dfrac6D = \dfrac{\log 5}{\log 2}\), so \(D = \dfrac{6\log 2}{\log 5} \approx 2.58\) hours.
  7. \(10^{(100 - 70)/10} = 10^3 = 1000\) times.
  8. Power law: \(\ln x^2 = \ln(3x + 4)\), so \(x^2 = 3x + 4\), so \((x - 4)(x + 1) = 0\). \(x = -1\) makes \(\ln x\) undefined: reject. Answer \(x = 4\) (check: \(2\ln 4 = \ln 16 = \ln(12 + 4)\) ✓).

Common mistake

Taking logs before isolating the power. From \(5 \cdot 3^x = 100\), writing \(x\log 15 = \log 100\) is wrong: the 5 does not belong inside the power. Divide first: \(3^x = 20\).

Skipping the check. The algebra of log equations can create answers that make a log of a negative number or of 0. They must be rejected, every time.

Where it leads