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AI HL · 12.6

Exponents & Logarithms

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12.6.2

Logarithms and solving exponential and log equations

12.6.2

Logarithms and solving exponential and log equations

Target: I can use log₁₀ and ln as the inverses of 10^x and e^x, and evaluate them with the calculator. (+6 more)

By the end of this section

Key points

Logarithms are inverses

  • \(\log_{10} b = x\) means \(b = 10^x\). \(\ln b = x\) means \(b = e^x\). The number e is about 2.718.
  • \(\log x\) and \(10^x\) undo each other. So do \(\ln x\) and \(e^x\). \(\ln(e^x) = x\) and \(e^{\ln x} = x\).
  • You can only take the log of a positive number. After you solve, check every answer in the original equation.
  • To solve a log equation, get the log on its own, then write it as a power. \(3 + 2\ln x = 1\) gives \(\ln x = -1\), so \(x = e^{-1}\).
  • On a log scale, each step of 1 multiplies the quantity by 10. A difference of \(\Delta\) on the scale means a ratio of \(10^{\Delta}\).

The laws of logarithms

  • \(\log(xy) = \log x + \log y\) and \(\log\dfrac{x}{y} = \log x - \log y\).
  • \(\log(x^m) = m\log x\). A root is a power: \(\log\sqrt{x} = \tfrac{1}{2}\log x\).
  • These work for any base, and for ln. \(\log_{10} 10 = 1\), \(\log_{10} 1000 = 3\), \(\ln e = 1\), \(\ln 1 = 0\).
  • \(\log x + \log y\) is not \(\log(x + y)\). There is no law for the log of a sum.

Solving equations with logs

  • Combine all the log terms into one log, then convert to a power. Multiply out the bracket before collecting x terms.
  • A constant is a log too: \(2 = \log_{10} 100\). So \(\log x + \log 6 = 2\) means \(6x = 100\).
  • Different bases, \(a^{f(x)} = b^{g(x)}\): take ln of both sides, bring the powers down, collect the x terms, factorise and divide.
  • \(\ln 5\) and \(\ln 3\) are just numbers. Give the exact answer in logs, then 3 significant figures.
  • Reject any answer that puts a negative number, or 0, inside a log.

Fitting a + b ln x

  • Put each data point into \(f(x)=a+b\ln x\). This gives two equations in a and b.
  • Subtract the equations: the a cancels, so \(b\,(\ln x_2-\ln x_1)=y_2-y_1\). Then \(b=\dfrac{y_2-y_1}{\ln(x_2/x_1)}\).
  • Put b back to find a. Keep the full values of a and b on the calculator until the last step.

On the calculator

Evaluating log and ln

  • Use the calculator for log of a number, and ln of a number, for example −log of 3 × 10⁻⁵.
  • Put a bracket round the whole argument of the log.
  • ln is base e. log with no base is base 10.

Screenshot and video to come

Solving an equation numerically

  • Use the solver to find x when the equation has x on both sides, for example 5^(x+2) = 3^(2x−1).
  • Give 3 significant figures unless the question asks for an exact answer.

Screenshot and video to come

Checking with a graph

  • Graph the left side and the right side as two functions.
  • The x-coordinate of the intersection is the solution. Compare it with your exact answer as a decimal.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Undoing e to the power x

\(\text{Solve } e^x = 0.2. \text{ Give x to 3 significant figures.}\)

Working
  1. Take ln of both sides

    \(\ln\left(e^x\right) = \ln 0.2\)

    ln undoes e^x. In general ln(e^x) = x, and the same for log and 10^x.

  2. Calculator

    From the calculator: \(x = \ln 0.2 = -1.61\)

    Give 3 significant figures.

Video walkthrough coming
Converting a log equation

\(\text{Find the exact value of } y \text{ if } \log_{10}(y-2) = -2.\)

Working
  1. Convert

    \(y - 2 = 10^{-2}\)

    log₁₀ b = x means b = 10^x.

  2. Evaluate

    \(y - 2 = \frac{1}{100}\)

    A negative index is a reciprocal.

  3. Solve

    \(y = 2.01\)

    Add 2 to both sides.

Video walkthrough coming
Isolate ln, then use e

\(\text{Solve } 3 + 2\ln x = 1. \text{ Give an exact answer, and x to 3 significant figures.}\)

Working
  1. Isolate ln x

    \(2\ln x = -2 \Rightarrow \ln x = -1\)

    Subtract 3, then divide by 2.

  2. Use e

    \(x = e^{-1}\)

    ln x = k means x = e^k. This is the exact answer.

  3. Decimal

    From the calculator: \(x = 0.368\)

    Give 3 significant figures.

Video walkthrough coming
A log in terms of p and q

\(\text{Let } p = \log_{10} a \text{ and } q = \log_{10} b. \text{ Write } \log_{10}\!\left(\frac{a^2\sqrt{b}}{1000}\right) \text{ in terms of p and q.}\) Find the constant term.

Working
  1. Split the fraction

    \(\log a^2 + \log\sqrt{b} - \log 1000\)

    Product law adds, quotient law subtracts.

  2. Bring down the powers

    \(2\log a + \tfrac{1}{2}\log b - \log 1000\)

    √b = b^(1/2). log₁₀ 1000 = 3, because 10³ = 1000.

  3. Substitute

    \(2p + \tfrac{1}{2}q - 3\)

    The constant term is −3.

Video walkthrough coming
Add logs, then convert

\(\text{Solve } \log_{10}(x+3) + \log_{10} 6 = 2. \text{ Give an exact answer.}\)

Working
  1. Combine

    \(\log_{10}\left(6(x+3)\right) = 2\)

    log a + log b = log(ab).

  2. Convert

    \(6(x+3) = 10^2 = 100\)

    log₁₀ b = 2 means b = 10².

  3. Solve

    \(x + 3 = \frac{50}{3} \Rightarrow x = \frac{41}{3}\)

    Check x + 3 > 0. It is.

Video walkthrough coming
Subtract logs, then use e

\(\text{Solve } \ln(x+6) - \ln(x-1) = 2. \text{ Give x to 3 significant figures.}\)

Working
  1. Combine

    \(\ln\frac{x+6}{x-1} = 2\)

    ln a − ln b = ln(a/b).

  2. Remove the ln

    \(\frac{x+6}{x-1} = e^2 \Rightarrow x + 6 = e^2(x-1)\)

    ln y = k means y = e^k. Multiply both sides by (x − 1).

  3. Collect x

    \(6 + e^2 = e^2x - x = x(e^2-1)\)

    Put all the x terms on one side and factorise.

  4. Solve

    \(x = \frac{6+e^2}{e^2-1}\) From the calculator: \(x = 2.10\)

    Check x > 1 so that both logs exist. It is.

Video walkthrough coming
Different bases: take ln

\(\text{Solve } 5^{x+2} = 3^{2x-1}. \text{ Give an exact answer, and x to 3 significant figures.}\)

Working
  1. Take ln

    \((x+2)\ln 5 = (2x-1)\ln 3\)

    ln(a^m) = m ln a brings the powers down. ln 5 and ln 3 are just numbers.

  2. Expand

    \(x\ln 5 + 2\ln 5 = 2x\ln 3 - \ln 3\)

    Multiply out both brackets.

  3. Collect x

    \(2\ln 5 + \ln 3 = x(2\ln 3 - \ln 5)\)

    x terms on one side, the rest on the other. Factorise.

  4. Divide

    \(x = \frac{2\ln 5 + \ln 3}{2\ln 3 - \ln 5}\) From the calculator: \(x = 7.35\)

    Most equations like this can also be solved by graphing both sides and finding the intersection.

Video walkthrough coming
Fitting a model a + b ln t

The population of a new town is modelled by \(P=a+b\ln t\) thousand people, where t is the time in years since the town was founded. The population is 2.1 thousand when \(t=2\) and 3.4 thousand when \(t=10\). Find b, then a. Find the population predicted when \(t=20\).

Working
  1. Two equations

    \(a+b\ln2=2.1\) and \(a+b\ln10=3.4\)

    Put each point in the model.

  2. Subtract

    \(b(\ln10-\ln2)=1.3\), so \(b=\dfrac{1.3}{\ln5}=0.808\)

    The a cancels. \(\ln10-\ln2=\ln5\).

  3. Find a

    \(a=2.1-0.808\ln2=1.54\)

    Put b into the first equation. Keep the full value of b on the calculator.

  4. Predict

    \(P=1.54+0.808\ln20=3.96\) thousand

    Time 20 years is outside the data, so the prediction is an estimate.

Video walkthrough coming
How many times greater?

The energy E, in joules, released by an earthquake of magnitude M satisfies \(\log_{10}E=4.8+1.5M\). How many times more energy does a magnitude 7 earthquake release than a magnitude 5 earthquake?

Working
  1. Difference in the log

    \(\log_{10}E_7-\log_{10}E_5=1.5\times7-1.5\times5=3\)

    The 4.8 cancels.

  2. Use the log law

    \(\log_{10}\dfrac{E_7}{E_5}=3\)

    A difference of logs is the log of the ratio.

  3. Convert

    \(\dfrac{E_7}{E_5}=10^{3}=1000\)

    A magnitude 7 earthquake releases 1000 times the energy.

Video walkthrough coming