1.5 Laws of exponents and introduction to logarithmsIB Maths: Applications and Interpretation HL: Revision notes

An exponent (index) tells you how many times to multiply the base. For integer exponents these laws hold, for $a\neq0$:

• $a^{m}\times a^{n}=a^{m+n}$, for example $5^{3}\times5^{-6}=5^{-3}$.
• $a^{m}\div a^{n}=a^{m-n}$, for example $6^{4}\div6^{3}=6$.
• $(a^{m})^{n}=a^{mn}$, for example $(2^{3})^{-4}=2^{-12}$.
• $(ab)^{n}=a^{n}b^{n}$, for example $(2x)^{4}=16x^{4}$.
• $a^{0}=1$ and $a^{-n}=\frac{1}{a^{n}}$. The laws only work when the bases are the same, or when you are raising a product to a power.

A negative exponent means reciprocal: $a^{-n}=\frac{1}{a^{n}}$, so $(2x)^{-3}=\frac{1}{(2x)^{3}}=\frac{1}{8x^{3}}$. Brackets decide what the exponent applies to: $(2x)^{-3}=\frac{1}{8x^{3}}$ but $2x^{-3}=\frac{2}{x^{3}}$. Worked example: simplify $\frac{(3x^{2})^{3}}{9x^{4}}$. $(3x^{2})^{3}=27x^{6}$, so $\frac{27x^{6}}{9x^{4}}=3x^{6-4}=3x^{2}$. A negative exponent never makes the number negative: $2^{-3}=\frac18$, not $-8$.

A logarithm answers the question 'what power of the base gives this number?'. The two statements below mean the same thing: $a^{x}=b\iff\log_{a}b=x.$ For example $2^{3}=8$ so $\log_{2}8=3$, and $10^{-3}=0.001$ so $\log_{10}0.001=-3$. Because a positive base raised to any power is positive, $b>0$: you cannot take the logarithm of zero or of a negative number. The base $a$ must be positive and not equal to $1$. Logarithms and exponentials are inverses: $\log_{a}\left(a^{x}\right)=x$ and $a^{\log_{a}b}=b$.

Two bases are used most often.

• Base 10: $\log_{10}x$, written $\lg x$ or $\log x$ on many calculators. Examples: $\log_{10}1000=3$, $\log_{10}0.01=-2$.
• Base $e$: $\log_{e}x$ is written $\ln x$, the natural logarithm. The number $e\approx2.718$ is the base of natural growth and decay. Useful results: $\log_{10}10^{x}=x$, $\ln e^{x}=x$, $e^{\ln x}=x$, $\ln e=1$, $\ln1=0$. Use your GDC to evaluate logarithms that are not exact: $\log_{10}250=2.40$ and $\ln250=5.52$ to 3 significant figures.

To solve $a^{x}=b$, write it as $x=\log_{a}b$ and evaluate with technology.

• $3^{x}=40$ gives $x=\log_{3}40=3.36$. If your GDC has no base-3 logarithm, use its equation solver on $3^{x}=40$.
• $10^{x}=50$ gives $x=\log_{10}50=1.70$.
• Base $e$: $80e^{-0.05t}=20$ gives $e^{-0.05t}=0.25$, so $-0.05t=\ln0.25$ and $t=27.7$. In context, a decay model $m=80e^{-0.05t}$ has initial value $80$ (when $t=0$), and the logarithm lets you find the time at which a given mass is reached.