IB · Unit 1 · 1C Logarithms
log and 10 undo each other. ln and e undo each other. A coefficient in the exponent becomes a power on the argument.
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Log undoes a power of 10
The formula is \(\log 10^k = k\).
\(\log 10^8 = 8\)
A power of 10 undoes a log
The formula is \(10^{\log a} = a\) for \(a > 0\).
\(10^{\log 19} = 19\)
The same pair with e and ln
\(\ln e^k = k\)
\(e^{\ln a} = a\) for \(a > 0\).
A coefficient in the exponent
\(n\log a = \log a^n\), so \(10^{n\log a} = a^n\).
\(10^{4\log 2} = 2^4 = 16\)
A negative coefficient gives a reciprocal: \(10^{-\log 5} = \dfrac{1}{5}\).
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.
One inverse pair. The exponent is already a single number or a single log.
\(\log 10^k = k\)
\(\log 10^8 = 8\)
Answer: \(8\)
\(\log 10^{12} = 12\)
Answer: \(12\)
\(\log 10^{-6} = -6\)
Answer: \(-6\)
\(\ln e^k = k\)
\(\ln e^{2.5} = 2.5\)
Answer: \(2.5\)
\(\ln e^{-4} = -4\)
Answer: \(-4\)
\(10^{\log a} = a\)
\(10^{\log 19} = 19\)
Answer: \(19\)
\(e^{\ln a} = a\)
\(e^{\ln 11} = 11\)
Answer: \(11\)
A coefficient in front of the log or ln. Move it onto the argument as a power, then invert.
\(n\log a = \log a^n\)
\(4\log 2 = \log 2^4 = \log 16\)
\(10^{\log 16} = 16\)
Answer: \(16\)
\(3\log 5 = \log 5^3 = \log 125\)
\(10^{\log 125} = 125\)
Answer: \(125\)
\(2\log 6 = \log 36\)
\(10^{\log 36} = 36\)
Answer: \(36\)
\(3\ln 2 = \ln 2^3 = \ln 8\)
\(e^{\ln 8} = 8\)
Answer: \(8\)
\(2\ln 7 = \ln 49\)
\(e^{\ln 49} = 49\)
Answer: \(49\)
\(4\log 3 = \log 81\)
\(10^{\log 81} = 81\)
Answer: \(81\)
\(5\ln 2 = \ln 32\)
\(e^{\ln 32} = 32\)
Answer: \(32\)
\(\dfrac{1}{2}\log 36 = \log 36^{\frac{1}{2}} = \log 6\)
\(10^{\log 6} = 6\)
Answer: \(6\)
A negative exponent. The inverse still applies, and the answer is a reciprocal or a unit fraction.
\(10^{-\log 5} = \dfrac{1}{10^{\log 5}}\)
\(10^{\log 5} = 5\)
\(\dfrac{1}{5}\)
Answer: \(\dfrac{1}{5}\)
\(10^{-\log 8} = \dfrac{1}{10^{\log 8}}\)
\(\dfrac{1}{8}\)
Answer: \(\dfrac{1}{8}\)
\(-2\log 3 = \log 3^{-2} = \log \dfrac{1}{9}\)
\(10^{\log \dfrac{1}{9}} = \dfrac{1}{9}\)
Answer: \(\dfrac{1}{9}\)
\(-3\ln 5 = \ln 5^{-3} = \ln \dfrac{1}{125}\)
\(e^{\ln \dfrac{1}{125}} = \dfrac{1}{125}\)
Answer: \(\dfrac{1}{125}\)
\(-2\ln 4 = \ln 4^{-2} = \ln \dfrac{1}{16}\)
\(\dfrac{1}{16}\)
Answer: \(\dfrac{1}{16}\)
\(-4\log 2 = \log 2^{-4} = \log \dfrac{1}{16}\)
\(\dfrac{1}{16}\)
Answer: \(\dfrac{1}{16}\)
\(e^{-\ln 9} = \dfrac{1}{e^{\ln 9}} = \dfrac{1}{9}\)
Answer: \(\dfrac{1}{9}\)
\(-4\ln 2 = \ln 2^{-4} = \ln \dfrac{1}{16}\)
\(\dfrac{1}{16}\)
Answer: \(\dfrac{1}{16}\)
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