This collection presents 20 step-by-step practice problems on the properties of powers, arranged in order of increasing difficulty. The aim is not merely to carry out computations, but to learn how to recognise the structure of an expression and determine which property of powers can be applied.
We shall study the product and quotient of powers with the same base, the power of a power, the power of a product, and the power of a quotient. We shall also work with zero, negative, and rational exponents.
In the first problems we shall apply one property at a time. In the later ones, however, we shall have to use several properties in succession, while paying careful attention to the order in which the operations are performed.
Whenever variables, denominators, or negative and rational exponents occur, we shall also examine the conditions under which the expressions are defined. In this way, each simplification will preserve not only the value of the expression, but also its original domain.
Exercise 1 — level ★☆☆☆☆
Evaluate:
\[ 2^3\cdot 2^4. \]
Answer
\[ 2^3\cdot 2^4=128. \]
Solution
The expression is a product of two powers with the same base, namely \(2\). When powers with the same base are multiplied, the base is kept and the exponents are added:
\[ a^m\cdot a^n=a^{m+n}. \]
Here the exponents are \(3\) and \(4\). Therefore
\[ 2^3\cdot 2^4=2^{3+4}=2^7. \]
By definition, \(2^7\) is the product of seven factors, each equal to \(2\):
\[ 2^7=2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2=128. \]
Hence
\[ 2^3\cdot 2^4=128. \]
Exercise 2 — level ★☆☆☆☆
Evaluate:
\[ \frac{5^6}{5^4}. \]
Answer
\[ \frac{5^6}{5^4}=25. \]
Solution
The numerator and denominator are powers with the same base, equal to \(5\). Since the base is non-zero, we may apply the property for the quotient of powers with the same base:
\[ \frac{a^m}{a^n}=a^{m-n}, \qquad a\neq0. \]
The exponent of the denominator must be subtracted from that of the numerator. Thus
\[ \frac{5^6}{5^4}=5^{6-4}=5^2. \]
We now evaluate the remaining power:
\[ 5^2=5\cdot5=25. \]
Therefore
\[ \frac{5^6}{5^4}=25. \]
Exercise 3 — level ★☆☆☆☆
Evaluate:
\[ \left(3^2\right)^3. \]
Answer
\[ \left(3^2\right)^3=729. \]
Solution
The expression consists of the power \(3^2\), which is itself raised to the third power. We are therefore dealing with a power of a power.
In this situation the exponents are not added. Instead, the base is kept and the exponents are multiplied:
\[ \left(a^m\right)^n=a^{mn}. \]
The inner exponent is \(2\), while the outer exponent is \(3\). Hence
\[ \left(3^2\right)^3=3^{2\cdot3}=3^6. \]
Evaluating the sixth power of \(3\), we obtain
\[ 3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729. \]
The same result could also be found by first computing \(3^2=9\) and then \(9^3=729\). Applying the property of a power of a power, however, makes the structure of the expression explicit. Therefore
\[ \left(3^2\right)^3=729. \]
Exercise 4 — level ★☆☆☆☆
Evaluate:
\[ (2\cdot5)^3. \]
Answer
\[ (2\cdot5)^3=1000. \]
Solution
The expression inside the brackets is the product \(2\cdot5\), and the whole product is raised to the third power. We may therefore apply the property for the power of a product:
\[ (ab)^n=a^n b^n. \]
This property tells us that the exponent must be applied to each factor of the product. Consequently,
\[ (2\cdot5)^3=2^3\cdot5^3. \]
We now evaluate the two powers separately:
\[ 2^3=8 \qquad\text{and}\qquad 5^3=125. \]
It follows that
\[ 2^3\cdot5^3=8\cdot125=1000. \]
We can verify the result by first evaluating the product inside the brackets:
\[ (2\cdot5)^3=10^3=1000. \]
Therefore
\[ (2\cdot5)^3=1000. \]
Exercise 5 — level ★☆☆☆☆
Evaluate:
\[ \left(\frac{6}{3}\right)^4. \]
Answer
\[ \left(\frac{6}{3}\right)^4=16. \]
Solution
The expression is a quotient raised to the fourth power. There are two correct ways to proceed: we may simplify the quotient first, or we may apply the property for the power of a quotient.
In this case the first method is the most immediate, because
\[ \frac{6}{3}=2. \]
Therefore
\[ \left(\frac{6}{3}\right)^4=2^4=16. \]
We can also obtain the same result by using the property
\[ \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}, \qquad b\neq0. \]
Since the denominator is \(3\neq0\), we may write
\[ \left(\frac{6}{3}\right)^4=\frac{6^4}{3^4}. \]
Evaluating the numerator and denominator gives
\[ 6^4=1296 \qquad\text{and}\qquad 3^4=81. \]
Hence
\[ \frac{6^4}{3^4}=\frac{1296}{81}=16. \]
Both methods therefore lead to the same conclusion:
\[ \left(\frac{6}{3}\right)^4=16. \]
Exercise 6 — level ★★☆☆☆
Simplify:
\[ x^4\cdot x^5. \]
Answer
\[ x^4\cdot x^5=x^9. \]
Solution
The expression is a product of two powers with the same base \(x\). The fact that the base is represented by a variable rather than by a specific number does not change the property that must be applied.
When two powers with the same base are multiplied, the base is kept and the exponents are added:
\[ a^m\cdot a^n=a^{m+n}. \]
Here the common base is \(x\), while the exponents are \(4\) and \(5\). Therefore
\[ x^4\cdot x^5=x^{4+5}=x^9. \]
No restriction on \(x\) is required, since only positive integer powers occur. Hence the identity holds for every \(x\in\mathbb{R}\), and we conclude that
\[ x^4\cdot x^5=x^9. \]
Exercise 7 — level ★★☆☆☆
Simplify:
\[ \frac{x^9}{x^4} \qquad (x\neq0). \]
Answer
\[ \frac{x^9}{x^4}=x^5 \qquad (x\neq0). \]
Solution
Before simplifying the expression, we must examine its domain. Since \(x^4\) appears in the denominator, the quotient is defined only when
\[ x^4\neq0. \]
A positive integer power of a real number is zero exactly when its base is zero. Therefore
\[ x^4\neq0 \iff x\neq0. \]
Under this condition, the numerator and denominator are powers with the same base \(x\). We may therefore apply the property
\[ \frac{a^m}{a^n}=a^{m-n}, \qquad a\neq0. \]
We keep the base \(x\) and subtract the exponent of the denominator from that of the numerator:
\[ \frac{x^9}{x^4}=x^{9-4}=x^5. \]
Although the simplified expression \(x^5\) is defined also at \(x=0\), the original quotient is not. The condition \(x\neq0\) must therefore be retained. Hence
\[ \frac{x^9}{x^4}=x^5 \qquad (x\neq0). \]
Exercise 8 — level ★★☆☆☆
Simplify:
\[ \left(x^3\right)^5. \]
Answer
\[ \left(x^3\right)^5=x^{15}. \]
Solution
The expression consists of the power \(x^3\), which is itself raised to the fifth power. We are therefore dealing with a power of a power.
In this case the exponents are not added. Instead, the base is kept and the inner and outer exponents are multiplied:
\[ \left(a^m\right)^n=a^{mn}. \]
Here the inner exponent is \(3\), while the outer exponent is \(5\). Consequently,
\[ \left(x^3\right)^5=x^{3\cdot5}=x^{15}. \]
Since all the exponents involved are positive integers, the identity is valid for every \(x\in\mathbb{R}\). Therefore
\[ \left(x^3\right)^5=x^{15}. \]
Exercise 9 — level ★★☆☆☆
Simplify:
\[ (3x)^3. \]
Answer
\[ (3x)^3=27x^3. \]
Solution
Inside the brackets we have the product of the numerical factor \(3\) and the variable \(x\). Since the whole product is raised to the third power, the exponent must be applied to both factors.
We use the property
\[ (ab)^n=a^n b^n. \]
Taking \(a=3\), \(b=x\), and \(n=3\), we obtain
\[ (3x)^3=3^3x^3. \]
We now evaluate the numerical power:
\[ 3^3=3\cdot3\cdot3=27. \]
Therefore
\[ (3x)^3=27x^3. \]
It would be incorrect to write \(3x^3\), because the exponent acts on the entire product enclosed in brackets, not only on the variable \(x\). Thus
\[ (3x)^3=27x^3. \]
Exercise 10 — level ★★☆☆☆
Simplify:
\[ \left(\frac{x}{2}\right)^4. \]
Answer
\[ \left(\frac{x}{2}\right)^4=\frac{x^4}{16}. \]
Solution
The whole quotient \(\displaystyle\frac{x}{2}\) is raised to the fourth power. Since the denominator is the non-zero number \(2\), the expression is defined for every \(x\in\mathbb{R}\).
We may therefore apply the property for the power of a quotient:
\[ \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}, \qquad b\neq0. \]
The exponent \(4\) must be applied both to the numerator and to the denominator. Hence
\[ \left(\frac{x}{2}\right)^4=\frac{x^4}{2^4}. \]
Since
\[ 2^4=16, \]
we obtain
\[ \left(\frac{x}{2}\right)^4=\frac{x^4}{16}. \]
Exercise 11 — level ★★☆☆☆
Evaluate:
\[ 4^0\cdot 7^2. \]
Answer
\[ 4^0\cdot 7^2=49. \]
Solution
The expression contains the power \(4^0\). To evaluate it, we use the definition of a power with zero exponent.
For every non-zero real number \(a\), we have
\[ a^0=1. \]
The condition \(a\neq0\) is essential, because within the algebraic theory of powers the expression \(0^0\) is left undefined.
In this exercise the base is \(4\), which is non-zero. Therefore
\[ 4^0=1. \]
We also have
\[ 7^2=7\cdot7=49. \]
Substituting these values into the original expression, we obtain
\[ 4^0\cdot7^2=1\cdot49=49. \]
Hence
\[ 4^0\cdot7^2=49. \]
Exercise 12 — level ★★☆☆☆
Evaluate:
\[ 3^{-2}. \]
Answer
\[ 3^{-2}=\frac{1}{9}. \]
Solution
The exponent is negative. The minus sign in the exponent does not determine the sign of the power; instead, it tells us to take the reciprocal of the corresponding power with positive exponent.
If \(a\neq0\) and \(n\in\mathbb{N}^*\), then
\[ a^{-n}=\frac{1}{a^n}. \]
The base in this exercise is \(3\), which is non-zero, so the power is well defined. Applying the definition with \(a=3\) and \(n=2\), we obtain
\[ 3^{-2}=\frac{1}{3^2}. \]
Since
\[ 3^2=9, \]
it follows that
\[ 3^{-2}=\frac{1}{9}. \]
It would be incorrect to write \(3^{-2}=-9\), because the minus sign belongs to the exponent and indicates a reciprocal; it does not indicate that the value of the power is negative.
Exercise 13 — level ★★★☆☆
Simplify:
\[ x^{-3}\cdot x^7 \qquad (x\neq0). \]
Answer
\[ x^{-3}\cdot x^7=x^4 \qquad (x\neq0). \]
Solution
The expression contains the power \(x^{-3}\). By definition,
\[ x^{-3}=\frac{1}{x^3}. \]
This reciprocal is defined only when \(x^3\neq0\), which is equivalent to
\[ x\neq0. \]
Under this condition, the two powers in the product have the same base \(x\). The property for the product of powers with the same base continues to hold for integer exponents:
\[ a^h\cdot a^k=a^{h+k}, \qquad a\neq0,\quad h,k\in\mathbb{Z}. \]
We therefore add the exponents algebraically, taking their signs into account:
\[ x^{-3}\cdot x^7=x^{-3+7}. \]
Since
\[ -3+7=4, \]
we obtain
\[ x^{-3}\cdot x^7=x^4. \]
The same result can be checked directly from the definition of a negative exponent:
\[ x^{-3}\cdot x^7=\frac{1}{x^3}\cdot x^7=\frac{x^7}{x^3}=x^{7-3}=x^4. \]
Although the simplified expression \(x^4\) is defined also at \(x=0\), the original expression is not. The condition \(x\neq0\) must therefore be retained. Hence
\[ x^{-3}\cdot x^7=x^4 \qquad (x\neq0). \]
Exercise 14 — level ★★★☆☆
Simplify:
\[ \left(x^2y^3\right)^4. \]
Answer
\[ \left(x^2y^3\right)^4=x^8y^{12}. \]
Solution
Inside the brackets we have the product of the two powers \(x^2\) and \(y^3\), and the whole product is raised to the fourth power.
We first apply the property for the power of a product:
\[ (ab)^n=a^n b^n. \]
Treating \(x^2\) and \(y^3\) as the two factors, we obtain
\[ \left(x^2y^3\right)^4=\left(x^2\right)^4\left(y^3\right)^4. \]
Each factor is now a power of a power. We therefore use
\[ \left(a^m\right)^n=a^{mn}. \]
Applying this property to the first factor gives
\[ \left(x^2\right)^4=x^{2\cdot4}=x^8. \]
Applying it to the second factor gives
\[ \left(y^3\right)^4=y^{3\cdot4}=y^{12}. \]
Substituting these results, we obtain
\[ \left(x^2y^3\right)^4=x^8y^{12}. \]
All the exponents involved are positive integers and no denominator occurs, so the identity holds for every \(x,y\in\mathbb{R}\).
Exercise 15 — level ★★★☆☆
Simplify:
\[ \left(\frac{2}{x}\right)^{-3} \qquad (x\neq0). \]
Answer
\[ \left(\frac{2}{x}\right)^{-3}=\frac{x^3}{8} \qquad (x\neq0). \]
Solution
Before simplifying the expression, we must examine its domain. The variable \(x\) appears in the denominator of the quotient \(\displaystyle\frac{2}{x}\), so we must require
\[ x\neq0. \]
Moreover, the whole quotient is raised to a negative exponent, so its base must also be non-zero. This second condition is automatically satisfied, because the numerator is \(2\neq0\), and therefore \(\displaystyle\frac{2}{x}\neq0\) whenever \(x\neq0\).
A negative exponent indicates the reciprocal of the corresponding power with positive exponent. For non-zero \(a\) and \(b\), we have
\[ \left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^n. \]
Applying this property with \(a=2\), \(b=x\), and \(n=3\), we obtain
\[ \left(\frac{2}{x}\right)^{-3}=\left(\frac{x}{2}\right)^3. \]
We now apply the property for the power of a quotient:
\[ \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}, \qquad b\neq0. \]
Hence
\[ \left(\frac{x}{2}\right)^3=\frac{x^3}{2^3}. \]
Since
\[ 2^3=8, \]
it follows that
\[ \left(\frac{2}{x}\right)^{-3}=\frac{x^3}{8}. \]
Even though \(x\) no longer appears in the denominator of the simplified expression, the restriction \(x\neq0\) must be retained because it belongs to the domain of the original expression. Therefore
\[ \left(\frac{2}{x}\right)^{-3}=\frac{x^3}{8} \qquad (x\neq0). \]
Exercise 16 — level ★★★☆☆
Evaluate:
\[ 25^{\frac{1}{2}}. \]
Answer
\[ 25^{\frac{1}{2}}=5. \]
Solution
The exponent is the rational number \(\displaystyle\frac{1}{2}\). A power with an exponent of the form \(\displaystyle\frac{1}{q}\) is interpreted by means of a root.
For \(a>0\) and \(q\in\mathbb{N}^*\), we define
\[ a^{\frac{1}{q}}=\sqrt[q]{a}. \]
When \(q=2\), this becomes
\[ a^{\frac{1}{2}}=\sqrt{a}. \]
Since the base \(25\) is positive, we may therefore write
\[ 25^{\frac{1}{2}}=\sqrt{25}. \]
The symbol \(\sqrt{25}\) denotes the principal square root of \(25\), that is, the unique non-negative number whose square is \(25\).
Since
\[ 5^2=25, \]
it follows that
\[ \sqrt{25}=5. \]
Therefore
\[ 25^{\frac{1}{2}}=5. \]
The answer is \(5\), not \(\pm5\). The two numbers \(5\) and \(-5\) are both solutions of the equation \(x^2=25\), but the radical \(\sqrt{25}\) denotes only the non-negative one.
Exercise 17 — level ★★★☆☆
Evaluate:
\[ 8^{\frac{1}{3}}. \]
Answer
\[ 8^{\frac{1}{3}}=2. \]
Solution
The exponent is \(\displaystyle\frac{1}{3}\). A power with this exponent represents the cube root of the base.
For a positive base \(a\), we have
\[ a^{\frac{1}{3}}=\sqrt[3]{a}. \]
Since \(8>0\), we may write
\[ 8^{\frac{1}{3}}=\sqrt[3]{8}. \]
We must therefore determine the number which, when raised to the third power, gives \(8\).
Since
\[ 2^3=2\cdot2\cdot2=8, \]
we obtain
\[ \sqrt[3]{8}=2. \]
Therefore
\[ 8^{\frac{1}{3}}=2. \]
Exercise 18 — level ★★★☆☆
Simplify:
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}} \qquad (x>0). \]
Answer
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}}=x \qquad (x>0). \]
Solution
The expression contains powers with rational exponents. Within the theory developed here, we assume that the base is strictly positive:
\[ x>0. \]
Under this condition, the usual properties of powers remain valid for rational exponents.
The two powers are multiplied together and have the same base \(x\). We may therefore use the property
\[ a^r\cdot a^s=a^{r+s}, \qquad a>0,\quad r,s\in\mathbb{Q}. \]
We keep the base \(x\) and add the exponents:
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}} =x^{\frac{2}{3}+\frac{1}{3}}. \]
The two fractions have the same denominator, so their numerators can be added:
\[ \frac{2}{3}+\frac{1}{3} =\frac{2+1}{3} =\frac{3}{3} =1. \]
Hence
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}}=x^1. \]
Since every number raised to the first power is equal to itself,
\[ x^1=x. \]
Therefore
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}}=x \qquad (x>0). \]
We can also verify the result by expressing the rational powers in terms of radicals:
\[ x^{\frac{2}{3}}=\left(\sqrt[3]{x}\right)^2 \qquad\text{and}\qquad x^{\frac{1}{3}}=\sqrt[3]{x}. \]
Consequently,
\[ x^{\frac{2}{3}}\cdot x^{\frac{1}{3}} =\left(\sqrt[3]{x}\right)^2\cdot\sqrt[3]{x} =\left(\sqrt[3]{x}\right)^3 =x. \]
Exercise 19 — level ★★★☆☆
Evaluate:
\[ 27^{\frac{2}{3}}. \]
Answer
\[ 27^{\frac{2}{3}}=9. \]
Solution
The exponent is the rational number \(\displaystyle\frac{2}{3}\). The denominator \(3\) indicates a cube root, while the numerator \(2\) indicates that the result of the root must be raised to the second power.
For \(a>0\), \(p\in\mathbb{Z}\), and \(q\in\mathbb{N}^*\), we have
\[ a^{\frac{p}{q}}=\left(\sqrt[q]{a}\right)^p. \]
In this exercise \(a=27\), \(p=2\), and \(q=3\). Therefore
\[ 27^{\frac{2}{3}}=\left(\sqrt[3]{27}\right)^2. \]
The cube root of \(27\) is \(3\), because
\[ 3^3=27. \]
Hence
\[ 27^{\frac{2}{3}}=3^2=9. \]
We can also use the equivalent representation
\[ a^{\frac{p}{q}}=\sqrt[q]{a^p}, \qquad a>0. \]
This gives
\[ 27^{\frac{2}{3}} =\sqrt[3]{27^2} =\sqrt[3]{729} =9, \]
since
\[ 9^3=729. \]
Both methods are correct. In this case, taking the cube root first is more convenient because it keeps the numbers involved smaller. Therefore
\[ 27^{\frac{2}{3}}=9. \]
Exercise 20 — level ★★★☆☆
Simplify:
\[ \left(x^{-2}\right)^3\cdot x^9 \qquad (x\neq0). \]
Answer
\[ \left(x^{-2}\right)^3\cdot x^9=x^3 \qquad (x\neq0). \]
Solution
The expression contains the power \(x^{-2}\). By definition,
\[ x^{-2}=\frac{1}{x^2}. \]
This expression is defined only when \(x\neq0\). The same restriction must therefore be retained throughout the simplification and in the final result.
We first simplify the power of a power \(\left(x^{-2}\right)^3\). For a non-zero base and integer exponents, we have
\[ \left(a^h\right)^k=a^{hk}. \]
The inner exponent is \(-2\), while the outer exponent is \(3\). Therefore
\[ \left(x^{-2}\right)^3=x^{(-2)\cdot3}=x^{-6}. \]
Substituting this result into the original expression gives
\[ \left(x^{-2}\right)^3\cdot x^9=x^{-6}\cdot x^9. \]
The two powers now have the same base \(x\). We may therefore add the exponents:
\[ x^{-6}\cdot x^9=x^{-6+9}. \]
Since
\[ -6+9=3, \]
we obtain
\[ \left(x^{-2}\right)^3\cdot x^9=x^3. \]
We can verify the result directly from the definition of a negative exponent:
\[ \left(x^{-2}\right)^3 =\left(\frac{1}{x^2}\right)^3 =\frac{1}{x^6}. \]
Hence
\[ \left(x^{-2}\right)^3\cdot x^9 =\frac{x^9}{x^6} =x^{9-6} =x^3. \]
Although the simplified expression \(x^3\) is defined at \(x=0\), the original expression is not. The condition \(x\neq0\) must therefore be preserved. Thus
\[ \left(x^{-2}\right)^3\cdot x^9=x^3 \qquad (x\neq0). \]
Exercise 21 — level ★★★★☆
Simplify:
\[ \frac{(4x^3)^2}{(2x^2)^3} \qquad (x\neq0). \]
Answer
\[ \frac{(4x^3)^2}{(2x^2)^3}=2 \qquad (x\neq0). \]
Solution
The variable \(x\) occurs in the denominator, so the expression is defined only when \(x\neq0\).
We expand the numerator and denominator separately, first using the power of a product and then the power of a power:
\[ (4x^3)^2=4^2\left(x^3\right)^2=16x^6 \]
and
\[ (2x^2)^3=2^3\left(x^2\right)^3=8x^6. \]
Substituting these results into the quotient gives
\[ \frac{(4x^3)^2}{(2x^2)^3} =\frac{16x^6}{8x^6} =\frac{16}{8}\cdot\frac{x^6}{x^6}. \]
Since \(x\neq0\), we have \(\displaystyle\frac{x^6}{x^6}=1\). Therefore
\[ \frac{(4x^3)^2}{(2x^2)^3}=2. \]
Exercise 22 — level ★★★★☆
Simplify:
\[ \frac{\left(a^2b^3\right)^4}{a^5b^7} \qquad (a\neq0,\ b\neq0). \]
Answer
\[ \frac{\left(a^2b^3\right)^4}{a^5b^7} =a^3b^5 \qquad (a\neq0,\ b\neq0). \]
Solution
The powers \(a^5\) and \(b^7\) occur in the denominator, so we must require \(a\neq0\) and \(b\neq0\).
We expand the numerator by applying the power of a product:
\[ \left(a^2b^3\right)^4 =\left(a^2\right)^4\left(b^3\right)^4 =a^{2\cdot4}b^{3\cdot4} =a^8b^{12}. \]
Substituting this result into the quotient, we obtain
\[ \frac{a^8b^{12}}{a^5b^7}. \]
For each base, we subtract the exponent in the denominator from the exponent in the numerator:
\[ \frac{a^8b^{12}}{a^5b^7} =a^{8-5}b^{12-7} =a^3b^5. \]
Therefore
\[ \frac{\left(a^2b^3\right)^4}{a^5b^7} =a^3b^5 \qquad (a\neq0,\ b\neq0). \]
Exercise 23 — level ★★★★☆
Simplify:
\[ \frac{(2x^3)^4}{(4x^2)^3} \qquad (x\neq0). \]
Answer
\[ \frac{(2x^3)^4}{(4x^2)^3} =\frac{x^6}{4} \qquad (x\neq0). \]
Solution
Since the variable \(x\) occurs in the denominator, the expression is defined only when \(x\neq0\).
We expand the numerator:
\[ (2x^3)^4 =2^4\left(x^3\right)^4 =16x^{12}. \]
We expand the denominator in the same way:
\[ (4x^2)^3 =4^3\left(x^2\right)^3 =64x^6. \]
Substituting the results gives
\[ \frac{(2x^3)^4}{(4x^2)^3} =\frac{16x^{12}}{64x^6} =\frac{16}{64}x^{12-6}. \]
Since \(\displaystyle\frac{16}{64}=\frac14\) and \(12-6=6\), it follows that
\[ \frac{(2x^3)^4}{(4x^2)^3} =\frac{x^6}{4} \qquad (x\neq0). \]
Exercise 24 — level ★★★★☆
Simplify:
\[ \left(a^{\frac12}b^{\frac13}\right)^6 \qquad (a\geq0,\ b\geq0). \]
Answer
\[ \left(a^{\frac12}b^{\frac13}\right)^6 =a^3b^2 \qquad (a\geq0,\ b\geq0). \]
Solution
The conditions \(a\geq0\) and \(b\geq0\) ensure that the rational powers occurring in the expression are defined within the framework being used.
We distribute the exponent \(6\) over the two factors:
\[ \left(a^{\frac12}b^{\frac13}\right)^6 =\left(a^{\frac12}\right)^6 \left(b^{\frac13}\right)^6. \]
We now apply the power-of-a-power property by multiplying the exponents:
\[ \left(a^{\frac12}\right)^6 =a^{\frac12\cdot6} =a^3 \]
and
\[ \left(b^{\frac13}\right)^6 =b^{\frac13\cdot6} =b^2. \]
Therefore
\[ \left(a^{\frac12}b^{\frac13}\right)^6=a^3b^2. \]
Exercise 25 — level ★★★★☆
Simplify:
\[ \left(\frac{x^2}{y^3}\right)^{-2} \qquad (x\neq0,\ y\neq0). \]
Answer
\[ \left(\frac{x^2}{y^3}\right)^{-2} =\frac{y^6}{x^4} \qquad (x\neq0,\ y\neq0). \]
Solution
The condition \(y\neq0\) is necessary for the quotient \(\displaystyle\frac{x^2}{y^3}\) to be defined. Since this quotient is raised to a negative exponent, its base must also be non-zero, so we must require \(x\neq0\) as well.
A quotient raised to a negative exponent is transformed into its reciprocal raised to the corresponding positive exponent:
\[ \left(\frac{x^2}{y^3}\right)^{-2} =\left(\frac{y^3}{x^2}\right)^2. \]
We now apply the exponent \(2\) to both numerator and denominator:
\[ \left(\frac{y^3}{x^2}\right)^2 =\frac{\left(y^3\right)^2}{\left(x^2\right)^2} =\frac{y^{3\cdot2}}{x^{2\cdot2}} =\frac{y^6}{x^4}. \]
Therefore
\[ \left(\frac{x^2}{y^3}\right)^{-2} =\frac{y^6}{x^4} \qquad (x\neq0,\ y\neq0). \]
Exercise 26 — level ★★★★☆
Simplify:
\[ \frac{2^n\cdot4^n}{8^n}. \]
Answer
\[ \frac{2^n\cdot4^n}{8^n}=1. \]
Solution
The bases \(2\), \(4\), and \(8\) are positive, so the powers are defined for every \(n\in\mathbb{R}\), and the denominator \(8^n\) is always non-zero.
We rewrite \(4\) and \(8\) as powers of \(2\):
\[ 4^n=\left(2^2\right)^n=2^{2n} \]
and
\[ 8^n=\left(2^3\right)^n=2^{3n}. \]
Substituting these expressions gives
\[ \frac{2^n\cdot4^n}{8^n} =\frac{2^n\cdot2^{2n}}{2^{3n}} =\frac{2^{n+2n}}{2^{3n}} =\frac{2^{3n}}{2^{3n}}. \]
Since \(2^{3n}\neq0\), the quotient equals \(1\). Therefore
\[ \frac{2^n\cdot4^n}{8^n}=1. \]
Exercise 27 — level ★★★★☆
Simplify:
\[ \frac{(3x^2)^3(2x)^2}{(6x^4)^2} \qquad (x\neq0). \]
Answer
\[ \frac{(3x^2)^3(2x)^2}{(6x^4)^2} =3 \qquad (x\neq0). \]
Solution
Since \(x\) occurs in the denominator, the expression is defined only when \(x\neq0\).
We expand the two factors in the numerator:
\[ (3x^2)^3 =3^3\left(x^2\right)^3 =27x^6 \]
and
\[ (2x)^2=2^2x^2=4x^2. \]
The numerator therefore becomes
\[ 27x^6\cdot4x^2=108x^8. \]
We now expand the denominator:
\[ (6x^4)^2=6^2\left(x^4\right)^2=36x^8. \]
Hence
\[ \frac{(3x^2)^3(2x)^2}{(6x^4)^2} =\frac{108x^8}{36x^8} =\frac{108}{36}\cdot\frac{x^8}{x^8} =3. \]
Exercise 28 — level ★★★★★
Simplify:
\[ \frac{a^{m+n}a^{m-n}}{\left(a^m\right)^2} \qquad (a\neq0). \]
Answer
\[ \frac{a^{m+n}a^{m-n}}{\left(a^m\right)^2}=1 \qquad (a\neq0). \]
Solution
The condition \(a\neq0\) ensures that the denominator is non-zero. If \(m\) and \(n\) are integers, all the powers involved are defined; more generally, the method applies whenever the powers are defined and the usual exponent laws hold.
The numerator contains two powers with the same base, so we add their exponents:
\[ a^{m+n}a^{m-n} =a^{(m+n)+(m-n)} =a^{2m}. \]
In the denominator, we apply the power-of-a-power property:
\[ \left(a^m\right)^2=a^{2m}. \]
The numerator and denominator are therefore equal. Hence
\[ \frac{a^{m+n}a^{m-n}}{\left(a^m\right)^2} =\frac{a^{2m}}{a^{2m}} =1. \]
The simplification relies on the fact that the exponents \(m+n\) and \(m-n\) have sum \(2m\).
Exercise 29 — level ★★★★★
Simplify:
\[ \frac{3^{n+2}-3^{n+1}}{2\cdot3^n}. \]
Answer
\[ \frac{3^{n+2}-3^{n+1}}{2\cdot3^n}=3. \]
Solution
Since the base \(3\) is positive, the powers are defined for every \(n\in\mathbb{R}\). Moreover, \(2\cdot3^n\neq0\), so the quotient is always defined.
We rewrite the two powers in the numerator by displaying the common factor \(3^n\):
\[ 3^{n+2}=3^n\cdot3^2=9\cdot3^n \]
and
\[ 3^{n+1}=3^n\cdot3=3\cdot3^n. \]
The numerator becomes
\[ 3^{n+2}-3^{n+1} =9\cdot3^n-3\cdot3^n =3^n(9-3) =6\cdot3^n. \]
Substituting into the quotient gives
\[ \frac{3^{n+2}-3^{n+1}}{2\cdot3^n} =\frac{6\cdot3^n}{2\cdot3^n} =\frac62 =3. \]
Exercise 30 — level ★★★★★
Simplify:
\[ \frac{x^{a+b}x^{b+c}x^{c+a}} {\left(x^ax^bx^c\right)^2} \qquad (x>0). \]
Answer
\[ \frac{x^{a+b}x^{b+c}x^{c+a}} {\left(x^ax^bx^c\right)^2} =1 \qquad (x>0). \]
Solution
The condition \(x>0\) ensures that all the powers are defined even when \(a\), \(b\), and \(c\) are real numbers. It also ensures that the denominator is non-zero.
In the numerator, we add the exponents of the powers with the same base:
\[ x^{a+b}x^{b+c}x^{c+a} =x^{(a+b)+(b+c)+(c+a)}. \]
The sum of the exponents is
\[ (a+b)+(b+c)+(c+a) =2a+2b+2c =2(a+b+c). \]
The numerator is therefore
\[ x^{2(a+b+c)}. \]
In the denominator, we first simplify the product inside the brackets:
\[ x^ax^bx^c=x^{a+b+c}. \]
Raising this result to the second power gives
\[ \left(x^{a+b+c}\right)^2=x^{2(a+b+c)}. \]
The numerator and denominator are equal. Therefore
\[ \frac{x^{a+b}x^{b+c}x^{c+a}} {\left(x^ax^bx^c\right)^2} =\frac{x^{2(a+b+c)}}{x^{2(a+b+c)}} =1. \]