Exercises — Laws of exponents — review with real exponents

WHAT: same base $2$, being multiplied — so we add exponents. WHY: a pile of five $2$'s next to a pile of three $2$'s is one pile of eight $2$'s (product law $a^m{a}^n=a^{m+n}$). $2^5\cdot 2^3=2^{5+3}=2^8.$ Answer: $2^8$ (which is $256$).

WHAT: same base, being divided — so we subtract exponents. WHY: dividing cancels copies; nine copies with four cancelled leaves five (quotient law $a^m/a^n=a^{m-n}$). $\frac{7^9}{7^4}=7^{9-4}=7^5.$ Answer: $7^5$.

WHAT: a minus on top means reciprocal (flip), not a negative number. WHY: $a^{-n}=1/a^n$ is forced by wanting $a^n{a}^{-n}=a^0=1$. $5^{-2}=\frac{1}{5^2}=\frac{1}{25}.$ Answer: $\frac{1}{25}$ — a positive number.

WHAT: a fraction on top = root then power. Denominator $3$ → cube root; numerator $2$ → square it. WHY: $a^{p/q}=(\sqrt[q]{a}{)}^p$, itself forced by $(a^{1/q}{)}^q=a$. $8^{2/3}=(\sqrt[3]{8}{)}^2=2^2=4.$ Answer: $4$.

WHAT: first the minus (flip), then the fraction (root & power). WHY: $a^{-p/q}=1/(\sqrt[q]{a}{)}^p$. $16^{-3/4}=\frac{1}{16^{3/4}}=\frac{1}{(\sqrt[4]{16}{)}^3}=\frac{1}{2^3}=\frac{1}{8}.$ Answer: $\frac{1}{8}$.

WHAT: split the fraction top and bottom, then root-and-power each piece. WHY: $(a/b{)}^n=a^n/b^n$; here $\sqrt[3]{27}=3$, $\sqrt[3]{125}=5$. ${\left(\frac{27}{125}\right)}^{2/3}=\frac{27^{2/3}}{125^{2/3}}=\frac{3^2}{5^2}=\frac{9}{25}.$ Answer: $\frac{9}{25}$.

Numerator: same base multiplied → add: $8^{2/3-1/3}=8^{1/3}$. Divide by $8^{-1/3}$: subtract exponents, and $-(-1/3)=+1/3$: $8^{1/3-(-1/3)}=8^{2/3}=(\sqrt[3]{8}{)}^2=4.$ Answer: $4$.

Power of a product: $(x^3{y}^{-2}{)}^2=x^6{y}^{-4}$ (multiply each exponent by $2$). Divide: subtract exponents base-by-base: $x^{6-4}{y}^{-4-(-5)}=x^2{y}^1=x^{2}y.$ Answer: $x^{2}y$.

WHAT: factor the common $2^n$ out of the top — there is no law for subtracting powers directly. WHY: $2^{n+3}=2^n\cdot 2^3$ by the product law, so the top is $2^n(2^3-1)$. $\frac{2^n(2^3-1)}{2^n}=2^3-1=8-1=7.$ Answer: $7$ (independent of $n$!).

WHAT: write both sides on the same base $2$, then match exponents. WHY: $4=2^2$ and $8=2^3$; since $a^x$ is one-to-one for $a>0,a\ne 1$, equal powers force equal exponents. $4^x=(2^2{)}^x=2^{2x}=2^3\Rightarrow 2x=3\Rightarrow x=\frac{3}{2}.$ Answer: $x=\frac{3}{2}$.

WHAT: add all the exponents — the product law survives for irrational exponents by the limit argument (see Limits and Continuity). $\sqrt{2}+\sqrt{2}-2\sqrt{2}+3=(2\sqrt{2}-2\sqrt{2})+3=3.$ The $\sqrt{2}$'s cancel like ordinary numbers. $=2^3=8.$ Answer: $8$.

WHAT: rewrite everything as a power of $3$. $9^x=(3^2{)}^x=3^{2x},27=3^3.$ Multiply on the left → add exponents: $3^{2x}\cdot 3^{x-1}=3^{3x-1}$. Match: $3x-1=3\Rightarrow 3x=4\Rightarrow x=\frac{4}{3}$. Answer: $x=\frac{4}{3}$.

Law says: $(a^m{)}^n=a^{mn}$ (multiply), so $(2^3{)}^2=2^6=64$. Tower would give: $2^{(3^2)}=2^9=512$. $64\ne 512$ — they are different. Conclusion: the true law is $==(a^m{)}^n=a^{mn}==$, and $a^{m^n}$ (a tower) is something else entirely. Notation without brackets is dangerous.

We want $a^{-n}$ to obey the product law together with $a^n$. Add the exponents: $a^n\cdot a^{-n}=a^{n+(-n)}=a^0=1.$ An equation $a^n\cdot X=1$ has exactly one solution: $X=\frac{1}{a^n}$. Therefore $a^{-n}=\frac{1}{a^n}$ — forced, not chosen. $■$

WHY it makes sense: $\sqrt{2}=1.41421\dots$ Each rational cut $1.4,1.41,1.414,\dots$ has a well-defined power (a root of $2$), and because $2^x$ is continuous and increasing for base $>1$, these powers close in on a single number we name $2^{\sqrt{2}}$ (see Number e and Natural Exponential for the same idea with base $e$). Squeeze: $2^{1.41}\approx 2.6574,2^{1.42}\approx \mathrm{2.6759.}$ The true value sits between, and $2^{1.414}\approx 2.6647$. Answer: $2^{\sqrt{2}}\approx 2.67$.