IB AA ยท 12B Logarithms
Rewrite the argument as a power of the base. The log is that exponent.
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\(\log_b(b^k)=k\)
\(\log A+\log B=\log(AB)\)
\(\ln a^n=n\ln a\)
\(\log_b a=\dfrac{\ln a}{\ln b}\)
Power of the base
The formula is \(\log_b(b^k)=k\).
32 = \(2^5\)
\(\log_2 32=5\)
Log of 1
The formula is \(\log_b 1=0\) because \(b^0=1\).
A unit fraction
Write the argument as a negative power.
\(\dfrac{1}{16}=2^{-4}\)
\(\log_2 \dfrac{1}{16}=-4\)
Change the base of the argument
Write both numbers as powers of the same prime.
\(27=3^3\) and \(9=3^2\)
\(\log_{27} 9=\dfrac{2}{3}\)
When a question writes \(\log\) with no base, that means \(\log_{10}\). \(\ln\) means \(\log_e\), the natural logarithm.
A different base is written as a subscript, for example \(\log_2 8\).
Rewrite as a power of the base, then read the exponent.
\(\log_b(b^k)=k\)
32 = \(2^5\)
\(\log_2 32=5\)
Answer: \(5\)
64 = \(2^6\)
\(\log_2 64=6\)
Answer: \(6\)
125 = \(5^3\)
\(\log_5 125=3\)
Answer: \(3\)
27 = \(3^3\)
\(\log_3 27=3\)
Answer: \(3\)
16 = \(4^2\)
\(\log_4 16=2\)
Answer: \(2\)
\(b^0=1\)
\(\log_6 1=0\)
Answer: \(0\)
The argument is 1 or a unit fraction.
\(\dfrac{1}{16}=2^{-4}\)
\(\log_2 \dfrac{1}{16}=-4\)
Answer: \(-4\)
\(\dfrac{1}{9}=3^{-2}\)
\(\log_3 \dfrac{1}{9}=-2\)
Answer: \(-2\)
\(\dfrac{1}{125}=5^{-3}\)
\(\log_5 \dfrac{1}{125}=-3\)
Answer: \(-3\)
\(\log_b 1=0\)
\(\log_4 1=0\)
Answer: \(0\)
\(\log_7 1=0\)
Answer: \(0\)
\(\dfrac{1}{32}=2^{-5}\)
\(\log_2 \dfrac{1}{32}=-5\)
Answer: \(-5\)
Write both numbers as powers of the same prime, then divide exponents.
\(27=3^3\)
\(9=3^2\)
\(\log_{27} 9=\dfrac{2}{3}\)
Answer: 23
\(8=2^3\)
\(2=2^1\)
\(\log_8 2=\dfrac{1}{3}\)
Answer: 13
\(32=2^5\)
\(8=2^3\)
\(\log_{32} 8=\dfrac{3}{5}\)
Answer: 35
\(81=3^4\)
\(\log_{81} 3=\dfrac{1}{4}\)
Answer: 14
\(16=2^4\)
\(8=2^3\)
\(\log_{16} 8=\dfrac{3}{4}\)
Answer: 34
\(27=3^3\)
\(81=3^4\)
\(\log_{27} 81=\dfrac{4}{3}\)
Answer: 43
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