IB · Unit 1 · 1C Logarithms
Use logs for pH, exponential growth and decay, and pairs of log equations. Give calculator answers to 3 significant figures.
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\(\log 10^k = k\)
\(10^{\log a} = a\)
\(\ln e^k = k\)
\(e^{\ln a} = a\)
pH
The formula is \(\mathrm{pH} = -\log[\mathrm{H}^+]\).
To go the other way, \([\mathrm{H}^+] = 10^{-\mathrm{pH}}\).
Initial value
In \(R = R_0 e^{kt}\), the initial value is at \(t = 0\).
\(e^{0} = 1\), so the initial value is \(R_0\).
Half-life
Set the model equal to half of the initial value, then take ln.
\(R_0 e^{kt} = \dfrac{1}{2} R_0\)
\(t = \dfrac{\ln \frac{1}{2}}{k}\)
Two log equations
\(\log(xy) = \log x + \log y\)
\(\log\dfrac{x}{y} = \log x - \log y\)
Add to find \(\log x\). Subtract to find \(\log y\).
When a question writes log with no base, that means log10.
ln means loge, the natural logarithm.
A different base is written as a subscript, for example log2 8. This is read as log base 2 of 8.
Exact answers stay in log or ln form, for example \(x = \log 6\) or \(x = \ln 5\). Do not switch to a decimal unless the question says to use technology.
Find a pH from a concentration, or read the initial value of an exponential model.
\(\mathrm{pH} = -\log[\mathrm{H}^+]\)
\(\mathrm{pH} = -\log(4.0 \times 10^{-6})\)
Use LOG on the calculator.
\(\mathrm{pH} = 5.40\) (3 s.f.)
Answer: \(5.40\)
\(\mathrm{pH} = -\log(3.2 \times 10^{-5})\)
\(\mathrm{pH} = 4.49\) (3 s.f.)
Answer: \(4.49\)
\(\mathrm{pH} = -\log(7.1 \times 10^{-9})\)
\(\mathrm{pH} = 8.15\) (3 s.f.)
Answer: \(8.15\)
Initial means t = 0.
\(e^{0} = 1\)
\(R = 80\)
Answer: \(80\)
\(t = 0\)
\(B = 500\)
Answer: \(500\)
\(\mathrm{pH} = -\log(6.3 \times 10^{-4})\)
\(\mathrm{pH} = 3.20\) (3 s.f.)
Answer: \(3.20\)
Undo pH to find a concentration, or solve an exponential model for the time.
\([\mathrm{H}^+] = 10^{-\mathrm{pH}}\)
\([\mathrm{H}^+] = 10^{-3.4}\)
\([\mathrm{H}^+] = 3.98 \times 10^{-4}\) (3 s.f.)
Answer: \(3.98 \times 10^{-4}\)
\([\mathrm{H}^+] = 10^{-8.2}\)
\([\mathrm{H}^+] = 6.31 \times 10^{-9}\) (3 s.f.)
Answer: \(6.31 \times 10^{-9}\)
\([\mathrm{H}^+] = 10^{-2.7}\)
\([\mathrm{H}^+] = 2.00 \times 10^{-3}\) (3 s.f.)
Answer: \(2.00 \times 10^{-3}\)
\(25 e^{-0.08t} = \dfrac{1}{2} \cdot 25\)
\(e^{-0.08t} = \dfrac{1}{2}\)
\(-0.08t = \ln \dfrac{1}{2}\)
\(t = \dfrac{\ln 2}{0.08}\)
\(t = 8.66\) days (3 s.f.)
Answer: \(\dfrac{\ln 2}{0.08}\) = 8.66 days
\(t = \dfrac{\ln 2}{0.12}\)
\(t = 5.78\) (3 s.f.)
Answer: \(5.78\)
\(1200 e^{0.15t} = 3600\)
\(e^{0.15t} = 3\)
\(t = \dfrac{\ln 3}{0.15}\)
\(t = 7.32\) (3 s.f.)
Answer: \(\dfrac{\ln 3}{0.15}\) = 7.32
Two log equations at once. Expand with the product and quotient laws, then add and subtract.
\(\log(xy) = \log x + \log y = 4\)
\(\log\dfrac{x}{y} = \log x - \log y = 2\)
Add the two equations.
\(2\log x = 6\)
\(\log x = 3\)
\(x = 10^3 = 1000\)
Subtract the two equations.
\(2\log y = 2\)
\(\log y = 1\)
\(y = 10\)
Answer: \(x = 1000,\ y = 10\)
\(\log x + \log y = 2\)
\(\log x - \log y = 0\)
\(2\log x = 2\)
\(x = 10\)
\(y = 10\)
Answer: \(x = 10,\ y = 10\)
\(2\log x = 6\)
\(x = 1000\)
\(2\log y = 4\)
\(y = 100\)
Answer: \(x = 1000,\ y = 100\)
\(2\log x = 0\)
\(x = 1\)
\(2\log y = 2\)
\(y = 10\)
Answer: \(x = 1,\ y = 10\)
\(2\log x = 4\)
\(x = 100\)
\(y = 100\)
Answer: \(x = 100,\ y = 100\)
\(2\log x = 2\)
\(x = 10\)
\(2\log y = 4\)
\(y = 100\)
Answer: \(x = 10,\ y = 100\)
\(2\log x = 8\)
\(x = 10^4 = 10\,000\)
\(2\log y = 4\)
\(y = 100\)
Answer: \(x = 10\,000,\ y = 100\)
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