Examples with solutions for Powers of a Fraction: Variable in the exponent of the power

Exercise #1

Insert the corresponding expression:

(23)a= \left(\frac{2}{3}\right)^a=

Video Solution

Step-by-Step Solution

Let's determine the corresponding expression for (23)a\left(\frac{2}{3}\right)^a:

We apply the property of exponentiation for fractions, which states:
(xy)n=xnyn\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}.

Substituting x=2x = 2, y=3y = 3, and n=an = a, we have:

(23)a=2a3a\left(\frac{2}{3}\right)^a = \frac{2^a}{3^a}.

Therefore, the correct expression is 2a3a \frac{2^a}{3^a} .

Assessing the possible choices:

  • Choice 1: 23a \frac{2}{3^a} - This is incorrect as it does not raise the numerator to aa.
  • Choice 2: 2a3a \frac{2a}{3a} - This is incorrect as it misuses the exponent rule.
  • Choice 3: 2a3a \frac{2^a}{3^a} - This is correct, as it follows the exponentiation property.
  • Choice 4: 2a3 \frac{2^a}{3} - This is incorrect as it does not raise the denominator to aa.

Thus, the correct choice is Choice 3: 2a3a \frac{2^a}{3^a} .

Answer

2a3a \frac{2^a}{3^a}

Exercise #2

Insert the corresponding expression:

(34)x= \left(\frac{3}{4}\right)^x=

Video Solution

Step-by-Step Solution

To solve this problem, we will apply the rules of exponents:

  • Step 1: Identify the Given Expression
  • Step 2: Apply the Exponent Rule

Now, let's work through these steps:

Step 1: We are given the expression (34)x\left(\frac{3}{4}\right)^x.

Step 2: According to the rule of exponents, when a fraction is raised to a power, this is equivalent to raising both the numerator and the denominator to that power. Therefore, we have:

(34)x=3x4x\left(\frac{3}{4}\right)^x = \frac{3^x}{4^x}

Therefore, the expression (34)x\left(\frac{3}{4}\right)^x is equivalent to 3x4x\frac{3^x}{4^x}.

Thus, the correct answer is option 1, which is 3x4x\frac{3^x}{4^x}.

The solution to the problem is 3x4x\frac{3^x}{4^x}.

Answer

3x4x \frac{3^x}{4^x}

Exercise #3

Insert the corresponding expression:

(57)ax= \left(\frac{5}{7}\right)^{ax}=

Video Solution

Step-by-Step Solution

To solve the problem, follow these steps:

  • Identify the given expression: (57)ax\left(\frac{5}{7}\right)^{ax}.
  • Apply the rule for exponentiation of a fraction: (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}.
  • Exponentiate both the numerator and the denominator separately by axax:
  • Calculate 5ax7ax\frac{5^{ax}}{7^{ax}} which follows directly from the application of the property.

Therefore, the rewritten expression is 5ax7ax\frac{5^{ax}}{7^{ax}}.

Among the given choices, the correct one is:

  • Choice 4: 5ax7ax \frac{5^{ax}}{7^{ax}}

Answer

5ax7ax \frac{5^{ax}}{7^{ax}}

Exercise #4

Insert the corresponding expression:

(59)2x+1= \left(\frac{5}{9}\right)^{2x+1}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll apply the power of a fraction rule, which states that (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}.

Step 1: Recognize the given expression: (59)2x+1\left(\frac{5}{9}\right)^{2x+1}.

Step 2: Apply the exponent rule to rewrite the expression. According to the rule, this becomes:

(59)2x+1=52x+192x+1\left(\frac{5}{9}\right)^{2x+1} = \frac{5^{2x+1}}{9^{2x+1}}.

Therefore, the expression can be rewritten as 52x+192x+1\frac{5^{2x+1}}{9^{2x+1}}.

Among the given choices, the correct option is choice 1: 52x+192x+1\frac{5^{2x+1}}{9^{2x+1}}.

Thus, the solution to the problem is 52x+192x+1 \frac{5^{2x+1}}{9^{2x+1}} .

Answer

52x+192x+1 \frac{5^{2x+1}}{9^{2x+1}}

Exercise #5

Insert the corresponding expression:

(1119)a+3b= \left(\frac{11}{19}\right)^{a+3b}=

Video Solution

Step-by-Step Solution

To solve this problem, we need to rewrite the expression (1119)a+3b\left(\frac{11}{19}\right)^{a+3b} using the rules for powers of fractions. Specifically, we apply the exponent to both the numerator and the denominator separately.

According to the rule (xy)n=xnyn\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}, we apply the exponent a+3ba+3b to both 11 and 19:

(1119)a+3b=11a+3b19a+3b \left(\frac{11}{19}\right)^{a+3b} = \frac{11^{a+3b}}{19^{a+3b}}

Therefore, the expression can be rewritten as 11a+3b19a+3b\frac{11^{a+3b}}{19^{a+3b}}, matching choice 3 in the provided options.

Hence, the solution to the problem is 11a+3b19a+3b\frac{11^{a+3b}}{19^{a+3b}}.

Answer

11a+3b19a+3b \frac{11^{a+3b}}{19^{a+3b}}

Exercise #6

Insert the corresponding expression:

(5×113×7)a= \left(\frac{5\times11}{3\times7}\right)^a=

Video Solution

Step-by-Step Solution

To solve this problem, follow these steps:

  • Identify the given expression: (5×113×7)a \left(\frac{5 \times 11}{3 \times 7}\right)^a .

  • Apply the exponent rule for powers of a fraction: (xy)a=xaya\left(\frac{x}{y}\right)^a = \frac{x^a}{y^a}.

  • Apply this rule separately to the numerator and the denominator:

The numerator is 5×115 \times 11 and the denominator is 3×73 \times 7. When the fraction is raised to a power aa, we apply the power to both the numerator and denominator:

(5×113×7)a=(5×11)a(3×7)a\left(\frac{5 \times 11}{3 \times 7}\right)^a = \frac{(5 \times 11)^a}{(3 \times 7)^a}

Which corresponds to option 1.

Each product is raised to the power aa. By exponent rules (xy)a=xa×ya(xy)^a = x^a \times y^a, this expression becomes:

5a×11a3a×7a\frac{5^a \times 11^a}{3^a \times 7^a}

Thus, the expression can be rewritten as: 5a×11a3a×7a\frac{5^a \times 11^a}{3^a \times 7^a}.

Referring to the provided choices, this matches choice 3.

Therefore, the correct choice is 4, A+C are correct.

Answer

A'+C' are correct

Exercise #7

Insert the corresponding expression:

(3×74×8)b+1= \left(\frac{3\times7}{4\times8}\right)^{b+1}=

Video Solution

Step-by-Step Solution

To solve the expression (3×74×8)b+1\left(\frac{3\times7}{4\times8}\right)^{b+1}, we follow these steps:

  • Step 1: Apply the power (b+1)(b+1) to the entire fraction, (3×74×8)b+1\left(\frac{3 \times 7}{4 \times 8}\right)^{b+1}.
  • Step 2: Use the exponentiation rule (ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n} to apply the power to both numerator and denominator separately.
  • Step 3: Expand the numerator and the denominator: (3×7)b+1=3b+1×7b+1 (3 \times 7)^{b+1} = 3^{b+1} \times 7^{b+1} and (4×8)b+1=4b+1×8b+1 (4 \times 8)^{b+1} = 4^{b+1} \times 8^{b+1} .
  • Step 4: Combine these results into one fraction: 3b+1×7b+14b+1×8b+1\frac{3^{b+1} \times 7^{b+1}}{4^{b+1} \times 8^{b+1}}.

Therefore, the simplified expression is 3b+1×7b+14b+1×8b+1 \frac{3^{b+1}\times7^{b+1}}{4^{b+1}\times8^{b+1}} .

Upon examining the choices, the correct option is choice 3: 3b+1×7b+14b+1×8b+1 \frac{3^{b+1}\times7^{b+1}}{4^{b+1}\times8^{b+1}} .

Answer

3b+1×7b+14b+1×8b+1 \frac{3^{b+1}\times7^{b+1}}{4^{b+1}\times8^{b+1}}

Exercise #8

Insert the corresponding expression:

(11×910×12)x+a= \left(\frac{11\times9}{10\times12}\right)^{x+a}=

Step-by-Step Solution

To solve the problem, we need to simplify the expression (11×910×12)x+a \left(\frac{11\times9}{10\times12}\right)^{x+a} and write it in the form requested in the question.

We begin by using the exponent rule: (ab)n=anbn (\frac{a}{b})^n = \frac{a^n}{b^n} . Applying this rule here:

<spanclass="katex">(11×910×12)x+a=(11×9)x+a(10×12)x+a</span><span class="katex"> \left(\frac{11\times9}{10\times12}\right)^{x+a} = \frac{(11\times9)^{x+a}}{(10\times12)^{x+a}} </span>

Next, we can simplify the expression further by applying the power over a product rule: (ab)n=an×bn (ab)^n = a^n \times b^n .

Applying this rule to both the numerator and denominator gives us:

Numerator: (11×9)x+a=11x+a×9x+a (11\times9)^{x+a} = 11^{x+a} \times 9^{x+a}

Denominator: (10×12)x+a=10x+a×12x+a (10\times12)^{x+a} = 10^{x+a} \times 12^{x+a}

Therefore, the entire expression becomes:

<spanclass="katex">11x+a×9x+a10x+a×12x+a</span><span class="katex"> \frac{11^{x+a} \times 9^{x+a}}{10^{x+a} \times 12^{x+a}} </span>

This matches the given answer. Thus, the solution to the question is:

11x+a×9x+a10x+a×12x+a \frac{11^{x+a}\times9^{x+a}}{10^{x+a}\times12^{x+a}}

Answer

11x+a×9x+a10x+a×12x+a \frac{11^{x+a}\times9^{x+a}}{10^{x+a}\times12^{x+a}}

Exercise #9

Insert the corresponding expression:

(32×5)x= \left(\frac{3}{2\times5}\right)^x=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the given expression.
  • Step 2: Apply the exponent rule (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} to distribute xx to numerator and denominator.
  • Step 3: Simplify the denominator expression by distributing exponent xx to each factor.

Now, let's work through each step:
Step 1: We have the original expression (32×5)x\left(\frac{3}{2 \times 5}\right)^x.
Step 2: Apply the rule to get 3x(2×5)x\frac{3^x}{(2 \times 5)^x}.
Step 3: Expand the denominator: (2×5)x=2x×5x(2 \times 5)^x = 2^x \times 5^x. This leads us to 3x2x×5x\frac{3^x}{2^x \times 5^x}.

Therefore, the solution to the problem is 3x2x×5x \frac{3^x}{2^x \times 5^x} .

Answer

3x2x×5x \frac{3^x}{2^x\times5^x}

Exercise #10

Insert the corresponding expression:

(137×6×3)x+y= \left(\frac{13}{7\times6\times3}\right)^{x+y}=

Video Solution

Step-by-Step Solution

Let's start by examining the expression given in the question:

(137×6×3)x+y \left(\frac{13}{7\times6\times3}\right)^{x+y}

This expression is a power of a fraction. There is a general rule in exponents which states:

(ab)n=anbn \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

Using this rule, we will apply it to our original expression.

Given, a=13 a = 13 , b=7×6×3 b = 7\times6\times3 , and n=x+y n = x+y , we can rewrite our expression as:

13x+y(7×6×3)x+y \frac{13^{x+y}}{(7\times6\times3)^{x+y}}

The solution to the question is:

13x+y(7×6×3)x+y \frac{13^{x+y}}{(7\times6\times3)^{x+y}}

Answer

13x+y(7×6×3)x+y \frac{13^{x+y}}{\left(7\times6\times3\right)^{x+y}}

Exercise #11

Insert the corresponding expression:

(23×5×7)x= \left(\frac{2}{3\times5\times7}\right)^x=

Video Solution

Step-by-Step Solution

To solve this problem, we need to express (23×5×7)x\left(\frac{2}{3 \times 5 \times 7}\right)^x by applying the rule for powers of a fraction.

Using the exponent rule (ab)x=axbx\left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}, we proceed as follows:

  • Step 1: Express the numerator and denominator with the exponent xx.
    The expression (23×5×7)x\left(\frac{2}{3 \times 5 \times 7}\right)^x becomes 2x(3×5×7)x\frac{2^x}{(3 \times 5 \times 7)^x}.
  • Step 2: Apply the power of a product rule to the denominator.
    This results in (3×5×7)x=3x×5x×7x(3 \times 5 \times 7)^x = 3^x \times 5^x \times 7^x.
  • Step 3: Substitute back into the fraction from Step 1.
    We get 2x3x×5x×7x\frac{2^x}{3^x \times 5^x \times 7^x}.

Therefore, the original expression (23×5×7)x\left(\frac{2}{3 \times 5 \times 7}\right)^x simplifies to 2x3x×5x×7x\frac{2^x}{3^x \times 5^x \times 7^x}.

The correct answer is: 2x3x×5x×7x \frac{2^x}{3^x \times 5^x \times 7^x} .

Answer

2x3x×5x×7x \frac{2^x}{3^x\times5^x\times7^x}

Exercise #12

Insert the corresponding expression:

(2×4×57)a= \left(\frac{2\times4\times5}{7}\right)^a=

Video Solution

Step-by-Step Solution

To solve this problem, we'll apply the exponent rule for fractions and products.

  • The given expression is (2×4×57)a \left(\frac{2 \times 4 \times 5}{7}\right)^a . We need to simplify this using exponent rules.
  • First, apply the exponent to both the numerator 2×4×52 \times 4 \times 5 and the denominator 77:
    (2×4×57)a=(2×4×5)a7a \left(\frac{2 \times 4 \times 5}{7}\right)^a = \frac{(2 \times 4 \times 5)^a}{7^a} .
  • Now, apply the rule (ab)n=an×bn (ab)^n = a^n \times b^n to distribute the exponent on the numerator:
  • (2×4×5)a=2a×4a×5a(2 \times 4 \times 5)^a = 2^a \times 4^a \times 5^a.
  • Therefore, the expression simplifies to:
    2a×4a×5a7a \frac{2^a \times 4^a \times 5^a}{7^a} .

Therefore, the expression simplifies to 2a×4a×5a7a \frac{2^a \times 4^a \times 5^a}{7^a} .

Answer

2a×4a×5a7a \frac{2^a\times4^a\times5^a}{7^a}

Exercise #13

Insert the corresponding expression:

(3×45×11×9)y= \left(\frac{3\times4}{5\times11\times9}\right)^y=

Video Solution

Step-by-Step Solution

To simplify the given expression, we start with the original expression:

(3×45×11×9)y \left(\frac{3\times4}{5\times11\times9}\right)^y .

Using the property for powers of a fraction, we distribute the exponent yy to the numerator and the denominator:

(ab)c=acbc \left(\frac{a}{b}\right)^c = \frac{a^c}{b^c}

First, apply the formula:

(3×45×11×9)y=(3×4)y(5×11×9)y \left(\frac{3\times4}{5\times11\times9}\right)^y = \frac{(3\times4)^y}{(5\times11\times9)^y} .

Next, apply the power of a product property, (ab)c=ac×bc(ab)^c = a^c \times b^c, to both the numerator and the denominator:

The numerator becomes (3×4)y=3y×4y(3\times4)^y = 3^y \times 4^y.

The denominator becomes (5×11×9)y=5y×11y×9y(5\times11\times9)^y = 5^y \times 11^y \times 9^y.

Thus, the fully simplified expression is:

3y×4y5y×11y×9y \frac{3^y\times4^y}{5^y\times11^y\times9^y} .

After comparing with the given options, this matches choice 1 and 2, so option 4 is the right one: A+B are correct

Answer

a'+b' are correct

Exercise #14

Insert the corresponding expression:

(2×7×215×6)x= \left(\frac{2\times7\times21}{5\times6}\right)^x=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Understand the expression and the exponent rules.
  • Step 2: Apply the exponent to both the numerator and denominator.
  • Step 3: Simplify each part of the expression by distributing the exponent correctly.

Now, let's work through each step:
Step 1: The original expression is (2×7×215×6)x\left(\frac{2\times7\times21}{5\times6}\right)^x. It needs to be rewritten by applying the exponent to each part of the fraction according to the rules of exponents.
Step 2: Using (ab)x=axbx\left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}, we apply the exponent xx to both the numerator and denominator:
(2×7×215×6)x=(2×7×21)x(5×6)x \left(\frac{2\times7\times21}{5\times6}\right)^x = \frac{(2\times7\times21)^x}{(5\times6)^x}
Step 3: Distribute the exponent xx across each multiplication in the numerator and the denominator according to the rule (a×b)x=ax×bx(a \times b)^x = a^x \times b^x:
In the numerator: (2×7×21)x=2x×7x×21x(2 \times 7 \times 21)^x = 2^x \times 7^x \times 21^x.
In the denominator: (5×6)x=5x×6x(5 \times 6)^x = 5^x \times 6^x.
Thus, the expression becomes:
2x×7x×21x5x×6x \frac{2^x \times 7^x \times 21^x}{5^x \times 6^x}

Therefore, the solution to the problem is 2x×7x×21x5x×6x\frac{2^x\times7^x\times21^x}{5^x\times6^x}.

Answer

2x×7x×21x5x×6x \frac{2^x\times7^x\times21^x}{5^x\times6^x}

Exercise #15

Insert the corresponding expression:

(5×6×79×11×13)b= \left(\frac{5\times6\times7}{9\times11\times13}\right)^b=

Video Solution

Step-by-Step Solution

To solve this problem, we'll use exponentiation properties to simplify the given expression:

  • Step 1: Apply the exponent to each term in the numerator: (5×6×7)b=5b×6b×7b(5 \times 6 \times 7)^b = 5^b \times 6^b \times 7^b.
  • Step 2: Apply the exponent to each term in the denominator: (9×11×13)b=9b×11b×13b(9 \times 11 \times 13)^b = 9^b \times 11^b \times 13^b.
  • Step 3: Use the property of exponents for fractions: (5×6×79×11×13)b=5b×6b×7b9b×11b×13b\left(\frac{5 \times 6 \times 7}{9 \times 11 \times 13}\right)^b = \frac{5^b \times 6^b \times 7^b}{9^b \times 11^b \times 13^b}.

As we have followed the rules of exponents and simplified accordingly, the corresponding expression is:

5b×6b×7b9b×11b×13b \frac{5^b \times 6^b \times 7^b}{9^b \times 11^b \times 13^b} .

Answer

5b×6b×7b9b×11b×13b \frac{5^b\times6^b\times7^b}{9^b\times11^b\times13^b}

Exercise #16

Insert the corresponding expression:

(2×4×67×8×9)3x= \left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}=

Video Solution

Step-by-Step Solution

Let's analyze the expression we are given:

(2×4×67×8×9)3x \left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}

The expression is a power of a fraction. The rule for powers of a fraction is that each component of the fraction must be raised to the power separately. This can be expressed as:

(ab)n=anbn \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

Applying this rule to our expression, we have:

  • The numerator inside the power: 2×4×6 2 \times 4 \times 6
  • The denominator inside the power: 7×8×9 7 \times 8 \times 9

Therefore, raising each part to the power 3x3x gives us:

(2×4×6)3x(7×8×9)3x \frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}

Thus, the simplified expression for the given equation is:

(2×4×6)3x(7×8×9)3x \frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}

The solution to the question is: (2×4×6)3x(7×8×9)3x \frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}

Answer

(2×4×6)3x(7×8×9)3x \frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}

Exercise #17

Insert the corresponding expression:

(23×7×5)a+2= \left(\frac{2}{3\times7\times5}\right)^{a+2}=

Video Solution

Step-by-Step Solution

To solve the problem, we will leverage the rules of exponents:

  • Apply the rule (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} to the expression (23×7×5)a+2\left(\frac{2}{3\times7\times5}\right)^{a+2}.
  • Distribute the exponent (a+2)(a+2) to both the numerator and the denominator.

First, distribute the exponent to the numerator:

2a+22^{a+2}

Now, distribute the exponent to the entire denominator:

(3×7×5)a+2(3\times7\times5)^{a+2}

Thus, the expression becomes:

2a+2(3×7×5)a+2\frac{2^{a+2}}{(3\times7\times5)^{a+2}}

This matches choice 3: 2a+2(3×7×5)a+2 \frac{2^{a+2}}{\left(3\times7\times5\right)^{a+2}} .

Answer

2a+2(3×7×5)a+2 \frac{2^{a+2}}{\left(3\times7\times5\right)^{a+2}}

Exercise #18

Insert the corresponding expression:

(5×6×79)2x+1= \left(\frac{5\times6\times7}{9}\right)^{2x+1}=

Video Solution

Step-by-Step Solution

Let's solve the problem step-by-step:

We begin with the expression: (5×6×79)2x+1 \left(\frac{5 \times 6 \times 7}{9}\right)^{2x+1} .

Step 1: Apply the exponent to both the numerator and the denominator using the rule (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}.

This gives: (5×6×7)2x+192x+1 \frac{(5 \times 6 \times 7)^{2x+1}}{9^{2x+1}} , as in choice b.

Step 2: Distribute the exponent across each factor in the numerator: (a×b×c)n=an×bn×cn (a \times b \times c)^n = a^n \times b^n \times c^n .

This results in: 52x+1×62x+1×72x+192x+1 \frac{5^{2x+1} \times 6^{2x+1} \times 7^{2x+1}}{9^{2x+1}} .

Therefore, the expression (5×6×79)2x+1\left(\frac{5\times6\times7}{9}\right)^{2x+1} evaluates to:

52x+1×62x+1×72x+192x+1 \frac{5^{2x+1} \times 6^{2x+1} \times 7^{2x+1}}{9^{2x+1}} .

This corresponds to choice 1. Hence, choices a' and b' are equivalent.

The correct answer is: a'+b' are correct.

Answer

a'+b' are correct

Exercise #19

Insert the corresponding expression:

(611×13×15)xy= \left(\frac{6}{11\times13\times15}\right)^{xy}=

Video Solution

Step-by-Step Solution

First, let's apply the exponent to the entire fraction:

(611×13×15)xy=6xy(11×13×15)xy \left(\frac{6}{11 \times 13 \times 15}\right)^{xy} = \frac{6^{xy}}{(11 \times 13 \times 15)^{xy}}

Now, distribute the xy xy exponent in the denominator to each factor:

6xy11xy×13xy×15xy \frac{6^{xy}}{11^{xy} \times 13^{xy} \times 15^{xy}}

Thus, the rewritten expression is 6xy11xy×13xy×15xy \frac{6^{xy}}{11^{xy} \times 13^{xy} \times 15^{xy}} .

Comparing our expression with the options given and based on our simplification, option 3: 6xy(11×13×15)xy \frac{6^{xy}}{\left(11 \times 13 \times 15\right)^{xy}} makes sense, as well as option 2: 6xy11xy×13xy×15xy \frac{6^{xy}}{11^{xy}\times13^{xy}\times15^{xy}} after distributing the exponent within the denominator.

Therefore, both options B and C are correct, making the right choice option 4.

Therefore, the correct answer is: B+C are correct.

Answer

B+C are correct

Exercise #20

Insert the corresponding expression:

1ax= \frac{1}{a^{-x}}=

Video Solution

Step-by-Step Solution

We begin with the expression: 1ax \frac{1}{a^{-x}} .
Our goal is to simplify this expression while converting any negative exponents into positive ones.

  • Recall the rule for negative exponents: an=1an a^{-n} = \frac{1}{a^n} .
  • Correspondingly, 1an=an \frac{1}{a^{-n}} = a^n .
  • Thus, in our expression 1ax \frac{1}{a^{-x}} , the negative exponent can be converted and flipped to the numerator by the rule: 1ax=ax \frac{1}{a^{-x}} = a^x .
Therefore, the expression evaluates to ax a^x .

The solution to the question is: ax a^x .

Answer

ax a^x