Laws of exponents — review with real exponents: Worked examples

Definition The domain rule you must never forget

Every law used on this page — $a^{x + y} = a^{x} a^{y}$ , $\frac{a ^{x}}{a ^{y}} = a^{x - y}$ , $(a^{x})^{y} = a^{x y}$ , the zero-exponent rule $a^{0} = 1$ , and $a^{p / q} = q a^{p}$ — is guaranteed to make sense in the real numbers only when the base $a > 0$ (or the exponent is a plain integer, where "copies" still works). We write $a > 0$ at the top of every problem in our heads. In the radical form $a^{p / q} = q a^{p}$ we require $p, q$ to be integers with $q$ a positive integer ( $q \geq 1$ ). That is what makes the principal-root convention well-posed: $q$ needs a whole-number index to name a single non-negative root. Two consequences we will see in action below:

Negative base + non-integer exponent = not real. $(- 2)^{1/2} = - 2$ and $(- 1)^{2}$ have no real value (Ex 5b). Never apply the laws to them.
Even-indexed radicals take the principal (non-negative) root. $481 = 3$ , not $\pm 3$ ; $q a$ with $q$ even means the one non-negative number whose $q$ -th power is $a$ . This convention is what keeps $a^{p / q}$ single-valued.

The scenario matrix

Every exponent expression you will ever meet lands in one of these cells. The right column names the example on this page that lives there.

#	Scenario class	Distinguishing feature	Covered by
A	Positive integer exponents	literal "copies"	Ex 1
B	Negative exponent	means reciprocal, not sign	Ex 2
C	Rational exponent (root + power)	$a^{p / q} = q a^{p}$ , principal root	Ex 2, Ex 3
D	Fraction base with $-$ fractional power	flip and root	Ex 3
E	Irrational exponent	limit-defined, still obeys algebra	Ex 4
F	Degenerate base: $a = 1$ , $a = 0$ (incl. $0^{0}$ , $0^{- n}$ )	edge behaviour	Ex 5
F $^{-}$	Forbidden base: $a < 0$ with non-integer exponent	not real	Ex 5b
G	Combining unlike bases	when the laws cannot merge	Ex 6
H	Real-world growth (word problem)	attach units, interpret	Ex 7
I	Exam twist / equation solving	match exponents	Ex 8

We work them in order. Read the "Forecast:" line and answer it in your head before unfolding the steps.

The examples

Worked example Example 2 — Cell B + C: negative and rational together

Simplify $1 6^{- 3/4}$ . (Base $16 > 0$ ✓, so the laws apply.)

Forecast: the minus sign is up top. Will the answer be negative, or a fraction? (Trap!)

Negative exponent means reciprocal: $1 6^{- 3/4} = \frac{1}{1 6 ^{3/4}}$ . Why this step? $a^{- n} = 1/ a^{n}$ is forced by wanting $a^{n} a^{- n} = a^{0} = 1$ . The minus never makes a number negative when $a > 0$ .
Split the rational power: $1 6^{3/4} = (1 6^{1/4})^{3} = (416)^{3}$ . Why this step? $a^{p / q} = (q a)^{p}$ — take the root first (smaller numbers), then the power. Here $416$ means the principal (positive) 4th root.
$416 = 2$ (since $2^{4} = 16$ ; we take $+ 2$ , not $- 2$ , by the principal-root convention), so $1 6^{3/4} = 2^{3} = 8$ .
Therefore $1 6^{- 3/4} = \frac{1}{8}$ .

Verify: $(\frac{1}{8})^{- 4/3}$ should give back $16$ : $8^{4/3} = (38)^{4} = 2^{4} = 16$ . ✓ And it's positive — the trap is defused. Answer: $\frac{1}{8}$ .

Worked example Example 3 — Cell C + D: fraction base, negative fractional power (geometric)

Simplify $(\frac{81}{16})^{- 3/4}$ . (Base $81/16 > 0$ ✓.)

Forecast: two things happen at once — a flip and a root. Which fraction ends up on top?

Negative flips the fraction: $(\frac{81}{16})^{- 3/4} = (\frac{16}{81})^{3/4}$ . Why this step? $a^{- n} = 1/ a^{n}$ applied to a whole fraction just turns it upside down.
Split top and bottom: $= \frac{1 6 ^{3/4}}{8 1 ^{3/4}}$ . Why this step? $(a / b)^{n} = a^{n} / b^{n}$ — the power distributes over division.
$1 6^{3/4} = (416)^{3} = 2^{3} = 8$ and $8 1^{3/4} = (481)^{3} = 3^{3} = 27$ . Note $481 = 3$ (principal root), not $\pm 3$ — the even-index radical returns the single non-negative value.
Answer $= \frac{8}{27}$ .

Look at the figure: the " $- 3/4$ " is really three separate instructions stacked on a number-line — flip, then 4th-root, then cube.

Verify: raise $\frac{8}{27}$ to $- 4/3$ : $(\frac{8}{27})^{- 4/3} = (\frac{27}{8})^{4/3} = \frac{( 3 27 ) ^{4}}{( 3 8 ) ^{4}} = \frac{3 ^{4}}{2 ^{4}} = \frac{81}{16}$ . ✓ Answer: $\frac{8}{27}$ .

Worked example Example 5 — Cell F: degenerate bases

$a = 1$ , $a = 0$ Evaluate (a) $1^{2}$ , (b) $1^{- 7}$ , (c) $0^{3}$ , (d) why $0^{0}$ is not defined here, (e) why $0^{- 2}$ is not defined.

Forecast: which of these are "danger" cases that have no single agreed value?

(a) $1^{2} = 1$ . Why this step? Multiplying any number of copies of $1$ (even a limit of them) is always $1$ ; $1^{x} = 1$ for every real $x$ .
(b) $1^{- 7} = \frac{1}{1 ^{7}} = \frac{1}{1} = 1$ . Same reason — reciprocal of $1$ is $1$ .
(c) $0^{3} = 0 \cdot 0 \cdot 0 = 0$ . For a positive exponent, piling copies of $0$ gives $0$ . (Our clean laws demanded $a > 0$ ; here we read $0^{n}$ directly by counting, valid only for $n > 0$ .)
(d) $0^{0}$ : the product law would need $0^{0} \cdot 0^{m} = 0^{0 + m} = 0^{m}$ for $m > 0$ , i.e. $0^{0} \cdot 0 = 0$ , which any value of $0^{0}$ satisfies — no unique answer. So it is left undefined in this real-exponent setting.
(e) $0^{- 2}$ : by the negative-exponent rule $0^{- 2} = \frac{1}{0 ^{2}} = \frac{1}{0}$ — division by zero, which is undefined. So $0$ raised to any negative power is undefined.

Verify: all three defined values plug back: $1^{2} = 1$ , $1^{- 7} = 1$ , $0^{3} = 0$ . ✓ Answer: $1, 1, 0$ ; and $0^{0}$ , $0^{- 2}$ both undefined.

Worked example Example 5b — Cell F

$^{-}$ : forbidden negative bases (this is why $a > 0$ ) Explain why $(- 2)^{1/2}$ and $(- 1)^{2}$ have no real value.

Forecast: the exponents look harmless. What breaks?

$(- 2)^{1/2} = - 2$ . Why this step? $a^{1/2}$ is defined as "the number whose square is $a$ ." But every real square is $\geq 0$ , so no real number squares to $- 2$ . Not real.
The double-value trap. If we tried to allow negative bases, $(- 8)^{2/6}$ could equal $6 (- 8)^{2} = 664 = + 2$ , yet the "cancelled" form $(- 8)^{1/3} = - 2$ . Same exponent, two answers — the laws would contradict each other.
$(- 1)^{2}$ . Why this step? For an irrational exponent we would need $lim (- 1)^{r}$ over rationals $r \to 2$ , but $(- 1)^{r}$ jumps between $+ 1$ , $- 1$ and non-real values as $r$ varies — no limit exists in the reals.
Conclusion. The laws on this page are all stated for $a > 0$ ; on a negative base with a non-integer exponent you must stop and refuse to apply them. (Complex analysis rescues these, but that is a different, multi-valued game.)

Verify: $(- 2)^{1/2}$ has no real value because $x^{2} = - 2$ has no real solution; $(- 1)^{2}$ has no real value because the required limit of $(- 1)^{r}$ does not exist. ✓ Answer: both are undefined over the real numbers.

Worked example Example 6 — Cell G: unlike bases you must NOT merge

Simplify as far as legally possible: $\frac{6 ^{5}}{2 ^{5} \cdot 3 ^{4}}$ . (All bases $> 0$ ✓.)

Forecast: the bases are $6, 2, 3$ — different! Can the laws touch this at all, or is it stuck?

Rewrite $6^{5} = (2 \cdot 3)^{5} = 2^{5} 3^{5}$ . Why this step? $(ab)^{n} = a^{n} b^{n}$ lets us split a product base — this is the only bridge between $6$ and the $2, 3$ downstairs.
Substitute: $\frac{2 ^{5} 3 ^{5}}{2 ^{5} 3 ^{4}}$ .
Cancel same-base factors: $\frac{2 ^{5}}{2 ^{5}} = 1$ (quotient law $2^{5 - 5} = 2^{0} = 1$ ) and $\frac{3 ^{5}}{3 ^{4}} = 3^{5 - 4} = 3^{1}$ . Why this step? Quotient law only fires between equal bases — that's why splitting $6$ in step 1 was essential.
Result $= 3$ .

Mistake watch: you could not write $\frac{6 ^{5}}{2 ^{5}}$ as $3^{5}$ and stop — you'd forget the leftover $3^{4}$ . Always split unlike bases into common ones first.

Verify: $6^{5} = 7776$ , $2^{5} \cdot 3^{4} = 32 \cdot 81 = 2592$ , and $7776/2592 = 3$ . ✓ Answer: $3$ .

Worked example Example 7 — Cell H: real-world growth (word problem, with units)

A colony of bacteria triples every $2$ hours. Starting mass is $50$ mg. Write the mass after $t$ hours and find the mass at $t = 5$ hours.

Forecast: at $t = 5$ hours (which is $2.5$ tripling-periods), is the mass more or less than $50 \times 3^{2} = 450$ mg?

Number of tripling periods in $t$ hours is $\frac{t}{2}$ , so mass $M (t) = 50 \cdot 3^{t /2}$ mg. Why this step? Each period multiplies by $3$ ; doing it $\frac{t}{2}$ times is exactly a real exponent $3^{t /2}$ — legal because the base $3 > 0$ (this is why real exponents matter, see Exponential Functions and their Graphs).
At $t = 5$ : $M (5) = 50 \cdot 3^{5/2} = 50 \cdot 3^{2} \cdot 3^{1/2} = 50 \cdot 9 \cdot 3$ . Why this step? Split $3^{5/2} = 3^{2} \cdot 3^{1/2}$ (product law) to separate the exact whole part from the root.
$= 4503 \approx 450 \times 1.732 \approx 779.4$ mg.

Look at the growth curve — $t = 5$ sits between the $t = 4$ point ( $450$ mg) and $t = 6$ point ( $1350$ mg), closer to being partway up, exactly as $3^{5/2}$ demands.

Verify (units + number): dimension is (mg)·(pure ratio) = mg ✓. Numerically $50 \cdot 3^{2.5} \approx 779.42$ mg, and indeed $450 < 779.4 < 1350$ . ✓ Answer: $M (t) = 50 \cdot 3^{t /2}$ mg; $M (5) = 4503 \approx 779.4$ mg.

Worked example Example 8 — Cell I: exam twist, solve for the exponent

Solve for $x$ : $(2)^{x} \cdot 4^{x} = 82$ . (All bases $> 0$ ✓.)

Forecast: every term is really a power of $2$ . Once rewritten, this becomes a plain linear equation — what will the coefficient of $x$ be?

Rewrite every base as a power of $2$ : $2 = 2^{1/2}$ , $4 = 2^{2}$ , $82 = 2^{3} \cdot 2^{1/2} = 2^{7/2}$ . Why this step? Matching one common base lets us compare exponents directly — the only way an exponential equation becomes solvable by algebra.
Left side: $(2^{1/2})^{x} \cdot (2^{2})^{x} = 2^{x /2} \cdot 2^{2 x} = 2^{x /2 + 2 x} = 2^{5 x /2}$ . Why this step? Power-of-a-power then product law: $\frac{x}{2} + 2 x = \frac{5 x}{2}$ .
Now $2^{5 x /2} = 2^{7/2}$ . Same base, so exponents must match: $\frac{5 x}{2} = \frac{7}{2}$ . Why this step? The function $a^{x}$ is monotonic (strictly one-directional) for $a > 0, a \neq = 1$ : if $a > 1$ it is strictly increasing, if $0 < a < 1$ it is strictly decreasing. Either way it never repeats a value, so it is one-to-one — equal powers therefore force equal exponents. Here $a = 2 > 1$ , strictly increasing.
$5 x = 7 \Rightarrow x = \frac{7}{5}$ .

Verify: put $x = \frac{7}{5}$ : exponent $\frac{5}{2} \cdot \frac{7}{5} = \frac{7}{2}$ , so left $= 2^{7/2} = 2^{3} 2 = 82$ . ✓ Answer: $x = \frac{7}{5}$ .

Active Recall

Recall What single condition on the base makes every law on this page legal?

$a > 0$ (or a plain integer exponent). It keeps $a^{x}$ single-valued, real, continuous, and monotonic.

Recall Why is

$(- 2)^{1/2}$ or $(- 1)^{2}$ forbidden? Negative base with a non-integer exponent has no real value: $- 2$ isn't real, and $(- 1)^{r}$ has no limit as $r \to 2$ . The laws simply don't apply — that's the whole reason for demanding $a > 0$ .

Recall What does

$481$ equal — $3$ or $\pm 3$ ? $3$ only. Even-indexed radicals return the principal (non-negative) root; this convention keeps $a^{p / q}$ single-valued.

Recall Which cell is the classic "sign trap"?

Cell B: $1 6^{- 3/4} = \frac{1}{8}$ is positive. The minus means reciprocal, never a negative value (for $a > 0$ ).

Recall Why is

$0^{- 2}$ undefined? $0^{- 2} = 1/ 0^{2} = 1/0$ — division by zero. $0$ to any negative power is undefined.

Recall Why is

$0^{0}$ left undefined here? The product law puts no unique constraint on it ( $0^{0} \cdot 0 = 0$ holds for any value), so no single answer survives.

Recall Strategy for "solve for the exponent" problems?

Rewrite every term as a power of one common base, combine with the laws, then set exponents equal (valid because $a^{x}$ is one-to-one — strictly increasing if $a > 1$ , strictly decreasing if $0 < a < 1$ ).

Connections

Rational Exponents and Radicals — the root-then-power moves and principal-root convention in Ex 2, 3.
Exponential Functions and their Graphs — the growth curve of Ex 7.
Limits and Continuity — why the irrational-exponent algebra in Ex 4 is legal, and why $(- 1)^{2}$ isn't.
Logarithms — Definition — the tidy tool to solve Ex 8 when bases don't match.
Binomial Theorem — needed the moment a sum base like $(a + b)^{n}$ appears.
Number e and Natural Exponential — the special base for continuous growth like Ex 7.