Question bank — Laws of exponents — review with real exponents

$2^3\cdot 2^4=4^7$ False. Same base, multiplying $\Rightarrow$ add exponents on the same base: $2^{3+4}=2^7$. The base stays $2$; it never doubles to $4$.

$a^m+a^n=a^{m+n}$ for all valid $a,m,n$ False. Exponents add only when the bases are multiplied, not added. $2^2+2^3=12$ but $2^5=32$; there is no simple law for a sum of powers.

$(a^m{)}^n=a^{m^n}$ False. The power-of-a-power law multiplies: $(a^m{)}^n=a^{mn}$. With $a=2,m=3,n=2$: $(2^3{)}^2=64=2^6$, but $2^{3^2}=2^9=512$. Towers are not the same as products.

$(a+b{)}^2=a^2+b^2$ False. Distribution works for products $(ab{)}^n=a^n{b}^n$, not sums. $(a+b{)}^2=a^2+2ab+b^2$ — use the Binomial Theorem. Test: $(1+1{)}^2=4\ne 1+1=2$.

$3^{-2}$ is a negative number False. A minus in the exponent means reciprocal, not sign: $3^{-2}=1/3^2=1/9>0$. For $a>0$ every real power $a^x$ is positive.

$2^{\sqrt{2}}\cdot 2^{\sqrt{2}}=2^{2\sqrt{2}}$ True. The product law survives for irrational exponents (it passes to the limit — see Limits and Continuity), so $\sqrt{2}+\sqrt{2}=2\sqrt{2}$ in the exponent.

$a^{2/6}=a^{1/3}$ for every real base $a$ False. True only for $a>0$. For $a<0$ the roots misbehave: $(-8{)}^{2/6}=\sqrt[6]{64}=+2$ but $(-8{)}^{1/3}=-2$ — same reduced fraction, different answers.

$a^0=1$ is just an arbitrary convention someone chose False. It is forced: the product law demands $a^m\cdot a^0=a^m$, so $a^0=a^m/a^m=1$. It is the only value that keeps the law consistent.

$(-1{)}^{1/2}$ is a perfectly good real number False. $\sqrt{-1}$ is not real, which is exactly why we restrict the base to $a>0$ for real-exponent laws.

$\frac{a^5}{a^5}=a^{5/5}=a^1=a$. Find the slip. They used the wrong law. Dividing same bases subtracts: $a^{5-5}=a^0=1$, not $a^{5/5}$. The exponent $5/5$ would come from a power/root, which isn't what division does.

$8^{-1/3}=-2$. Find the slip. They read the minus as a sign. It means reciprocal: $8^{-1/3}=1/8^{1/3}=1/2$. The value is a positive fraction, not $-2$.

${\left(\frac{a}{b}\right)}^{-2}=\frac{a^{-2}}{b^{-2}}=\frac{b^2}{a^2}$ but then claims this equals $\frac{a^2}{b^2}$. Find the slip. The first two steps are correct; the final claim is wrong. $\frac{b^2}{a^2}$ is the reciprocal of $\frac{a^2}{b^2}$ — the negative exponent flips the whole fraction, so the answer is $b^2/a^2$.

$2^3+2^3=2^6$. Find the slip. Adding equal powers is not the product law. $2^3+2^3=2\cdot 2^3=2^4=16$, whereas $2^6=64$. The correct move factors: $a^m+a^m=2a^m$.

$(2^{1/2}{)}^2=2^{1/2+2}=2^{5/2}$. Find the slip. They added instead of multiplied. Power-of-a-power multiplies exponents: $(2^{1/2}{)}^2=2^{(1/2)\cdot 2}=2^1=2$, which is exactly why $2^{1/2}=\sqrt{2}$.

$x^{-n}=\frac{1}{x^{-n}}$. Find the slip. They flipped a sign that was already the flip. $x^{-n}=\frac{1}{x^n}$ (positive $n$ downstairs). Writing $x^{-n}$ again downstairs undoes nothing and is circular.

$a>0$ demanded for real exponents, but $(-8{)}^{1/3}=-2$ still "works"? Isolated integer-index roots like cube roots of negatives can be defined, but the laws break: reducing $1/3$ to $2/6$ then gives $+2$. To keep $a^x$ single-valued, continuous and monotonic for all real $x$, we need $a>0$.

$a^{-n}$ mean "reciprocal" rather than "negative"? It is forced by wanting the product law to hold: $a^n\cdot a^{-n}=a^{n+(-n)}=a^0=1$, so $a^{-n}$ must be the multiplicative inverse $1/a^n$.

$a^{1/q}=\sqrt[q]{a}$, where $q$ is a positive whole number — why can't we pick another value? The power law forces $(a^{1/q}{)}^q=a^{(1/q)\cdot q}=a^1=a$. So $a^{1/q}$ is the number whose $q$-th power is $a$, which is exactly the $q$-th root $\sqrt[q]{a}$. See Rational Exponents and Radicals.

$a^{\sqrt{2}}$ when "$\sqrt{2}$ copies of $a$" is meaningless? Look at Figure 1. Along the horizontal axis we squeeze the irrational $\sqrt{2}$ between the rational numbers $r=1.4,1.41,1.414,\dots$ (here $r$ just means "a fraction we can already handle"). Each $a^r$ (blue dots) is defined by the rational-exponent rules. Because $a^x$ is continuous and monotonic for $a>0$, those dots march steadily toward one height (the red mark) — we name that height $a^{\sqrt{2}}$. "Copies" was only ever the integer starting point.

Sketch of the limit argument (see Figure 1 again). Take the product law $a^r{a}^s=a^{r+s}$, true for every rational pair $r,s$. Now let $r\to \sqrt{2}$ and $s\to \sqrt{3}$ along fractions. The left side is a product of two convergent sequences, so it converges to $a^{\sqrt{2}}\cdot a^{\sqrt{3}}$ (a limit of a product is the product of the limits). The right side has exponent $r+s\to \sqrt{2}+\sqrt{3}$, so it converges to $a^{\sqrt{2}+\sqrt{3}}$. Because both sides were equal at every stage, their limits are equal: $a^{\sqrt{2}}{a}^{\sqrt{3}}=a^{\sqrt{2}+\sqrt{3}}$. That is what "limits respect $+$ and $\times$" means. See Limits and Continuity.

$1^x$ for any real $x$ (even $x=\sqrt{2}$ or $x=-\pi$)? Always $1$. Every power of $1$ is $1$: $1^{\sqrt{2}}=1$, $1^{-\pi }=1/1^{\pi }=1$. The base $1$ is a fixed point of exponentiation.

$1^0$ — the rule $1^x=1$ or the rule $a^0=1$? No conflict: $a=1$ gives $1$ and $x=0$ gives $1$, so $1^0=1$ both ways. (It is $0^0$, not $1^0$, that is genuinely ambiguous.)

$0^0$? Reason it through both directions before deciding. It is indeterminate. Push it two ways. Direction A: for any $x>0$, $0^x=0$ (zero times itself is zero), so approaching along the base "$0$" suggests the answer should be $0$. Direction B: for any $a>0$, $a^0=1$ (forced by the product law), so approaching along the exponent "$0$" suggests $1$. Two honest limits, two different answers — so no single value is forced. In algebra/combinatorics people often define $0^0=1$ for convenience, but as a limit it has no fixed value.

$0^{-1}$ defined? No. $0^{-1}=1/0^1=1/0$, division by zero. Negative and zero exponents on base $0$ are undefined, which is one reason the clean laws assume $a>0$ strictly.

$0<a<1$, does raising to a larger exponent give a larger number? Explain why. No — it gives a smaller one. Why: for $0<a<1$, multiplying by $a$ shrinks whatever you have (e.g. $\times \frac{1}{2}$ halves it). So each extra factor of $a$ in $a^x$ shrinks the running total, meaning bigger $x$ = more shrinking = smaller value. Concretely $(1/2{)}^3=\frac{1}{8}<(1/2{)}^2=\frac{1}{4}$. Figure 2 shows this: the curve for $a=1/2$ (orange) slides downhill as $x$ grows, while $a=2$ (blue) climbs. So $a^x$ is decreasing when $0<a<1$ and increasing when $a>1$. See Exponential Functions and their Graphs.

$a=e$ obey different exponent laws from any other base? No. $e$ is a specific positive constant $\approx 2.718$, so $e^x$ follows the same laws. Its specialness is about calculus, not algebra — see Number e and Natural Exponential.

$a^{m-n}$ ever undefined when $m<n$ (so the exponent is negative)? Not for $a>0$. A negative exponent just means reciprocal: $a^{m-n}=1/a^{n-m}>0$, always finite. Trouble only appears at $a=0$.