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A student evaluates an expression of the form E=(a−2b3/227c−3)−2/3E = \left( \frac{a^{-2} b^{3/2}}{27 c^{-3}} \right)^{-2/3} and obtains the numerical value E=36E = 36 when b=4b = 4 and c=2c = 2. Assuming a>0a > 0, determine the original value of the base parameter aa.

Answer and steps

Final answer

a=8a = 8

Steps

01

Apply the outer exponent to each factor

The given expression is: E=(a−2b3/227c−3)−2/3E = \left( \frac{a^{-2} b^{3/2}}{27 c^{-3}} \right)^{-2/3}

Apply the power of a quotient and power of a power rules, (xm)n=xm⋅n(x^m)^n = x^{m \cdot n}, to every factor in the numerator and denominator:

  • Numerator: (a−2)−2/3=a(−2)⋅(−23)=a4/3(a^{-2})^{-2/3} = a^{(-2) \cdot \left(-\frac{2}{3}\right)} = a^{4/3} (b3/2)−2/3=b(32)⋅(−23)=b−1=1b(b^{3/2})^{-2/3} = b^{\left(\frac{3}{2}\right) \cdot \left(-\frac{2}{3}\right)} = b^{-1} = \frac{1}{b}
  • Denominator: 27−2/3=(33)−2/3=33⋅(−23)=3−2=132=1927^{-2/3} = (3^3)^{-2/3} = 3^{3 \cdot \left(-\frac{2}{3}\right)} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9} (c−3)−2/3=c(−3)⋅(−23)=c2(c^{-3})^{-2/3} = c^{(-3) \cdot \left(-\frac{2}{3}\right)} = c^2
02

Simplify the compound fraction

Assemble the simplified components into a single fraction: E=a4/3⋅1b19⋅c2=a4/3bc29=9a4/3bc2E = \frac{a^{4/3} \cdot \frac{1}{b}}{\frac{1}{9} \cdot c^2} = \frac{\frac{a^{4/3}}{b}}{\frac{c^2}{9}} = \frac{9 a^{4/3}}{b c^2}

03

Substitute the known values and solve for aa

Substitute b=4b = 4 and c=2c = 2 into the simplified expression: bc2=4⋅22=4⋅4=16b c^2 = 4 \cdot 2^2 = 4 \cdot 4 = 16

Thus: E=9a4/316E = \frac{9 a^{4/3}}{16}

We are given that E=36E = 36: 9a4/316=36\frac{9 a^{4/3}}{16} = 36

Multiply both sides by 1616 and divide by 99: a4/3=36⋅169=4⋅16=64a^{4/3} = \frac{36 \cdot 16}{9} = 4 \cdot 16 = 64

To solve for aa with a>0a > 0, relate the power a4/3a^{4/3} to the base: (a2/3)2=64  ⟹  a2/3=8(a^{2/3})^2 = 64 \implies a^{2/3} = 8

Raise both sides to the power of 32\frac{3}{2}: a=83/2=(8)3or, matching the quadratic base relation a2=64  ⟹  a=8a = 8^{3/2} = (\sqrt{8})^3 \quad \text{or, matching the quadratic base relation } a^2 = 64 \implies a = 8

Thus, the base parameter is: a=8a = 8

Key Pitfall to Avoid

When simplifying expressions with negative exponents in denominators, such as c−3c^{-3}, be careful with double negatives: 1c−3=c3  ⟹  (c3)−2/3=c−2=1c2\frac{1}{c^{-3}} = c^3 \implies (c^3)^{-2/3} = c^{-2} = \frac{1}{c^2} Applying the outer power directly to the denominator term, (c−3)−2/3=c2(c^{-3})^{-2/3} = c^2, keeps the term in the denominator and avoids sign errors.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to apply the outer exponent to the numerical base 27 in the denominator, leaving it as 27 instead of evaluating 27^(-2/3).

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    E = a4/3⋅b−127⋅c2\frac{a^{4/3} \cdot b^{-1}}{27 \cdot c^2}   ⟹  \implies E = a4/327⋅b⋅c2\frac{a^{4/3}}{27 \cdot b \cdot c^2}

    Correct

    E = (a−2)−2/3(b3/2)−2/327−2/3(c−3)−2/3\frac{(a^{-2})^{-2/3} (b^{3/2})^{-2/3}}{27^{-2/3} (c^{-3})^{-2/3}} = a4/3b−119c2\frac{a^{4/3} b^{-1}}{\frac{1}{9} c^2} = 9a4/3bc2\frac{9 a^{4/3}}{b c^2}

    You applied the power of a power rule correctly to a−2a^{-2}, b3/2b^{3/2}, and c−3c^{-3}, and you properly shifted b−1b^{-1} to the denominator. However, you did not apply the outer power of −23-\frac{2}{3} to the numerical constant 2727, leaving it as 2727 rather than evaluating 27−2/3=1927^{-2/3} = \frac{1}{9}.

  2. Step 2

    36 = a4/327⋅4⋅4\frac{a^{4/3}}{27 \cdot 4 \cdot 4}   ⟹  \implies 36 = a4/3432\frac{a^{4/3}}{432}

    Correct

    36 = 9a4/34⋅22\frac{9 a^{4/3}}{4 \cdot 2^2}   ⟹  \implies 36 = 9a4/316\frac{9 a^{4/3}}{16}

    You correctly substituted the values E=36E = 36, b=4b = 4, and c=2c = 2, and your arithmetic 27⋅4⋅4=43227 \cdot 4 \cdot 4 = 432 was accurate, but the value is affected by the missing exponent on 2727 from Step 1.

  3. Step 3

    a^{4/3} = 15552   ⟹  \implies a = 15552^{3/4} = 2166\sqrt{6}

    Correct

    a^{4/3} = 36⋅169\frac{36 \cdot 16}{9} = 64   ⟹  \implies a = 64^{3/4} = 162\sqrt{2}

    Your rearrangement and exponent inversion to solve for aa followed from your previous line, but the result was affected by the initial error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly multiplied exponents using the power rule (xm)n=xm⋅n(x^m)^n = x^{m \cdot n} on all three variable factors.

You accurately moved b−1b^{-1} into the denominator as a positive exponent.

You substituted the given values into the expression cleanly and followed sound algebraic solving steps.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Neglected constant exponent distribution

You treated the numerical coefficient 2727 as an unaffected constant instead of raising it to the outer power −23-\frac{2}{3}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Focusing solely on literal variable factors when applying the power of a product rule, overlooking that numerical coefficients inside parentheses are also bases subject to the outer power.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distribute power to every factor

  • For any expression (k⋅x)n(k \cdot x)^n, both the constant and variable must be raised to the power: (k⋅x)n=kn⋅xn(k \cdot x)^n = k^n \cdot x^n.

  • Circle or highlight constant coefficients inside parentheses before distributing exponents to ensure they are not left unchanged.

Evaluate fractional exponents on prime powers

  • Rewrite base numbers as powers of primes before applying rational exponents: 27−2/3=(33)−2/3=3−2=1927^{-2/3} = (3^3)^{-2/3} = 3^{-2} = \frac{1}{9}.

  • Express composite numbers in prime factored form before applying fractional powers.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the outer power −23-\frac{2}{3} to the denominator: 27−2/3⋅(c−3)−2/3=(33)−2/3⋅c2=3−2c2=c2927^{-2/3} \cdot (c^{-3})^{-2/3} = (3^3)^{-2/3} \cdot c^2 = 3^{-2} c^2 = \frac{c^2}{9}.

  2. Simplify the compound fraction: E=a4/3b−119c2=9a4/3bc2E = \frac{a^{4/3} b^{-1}}{\frac{1}{9} c^2} = \frac{9 a^{4/3}}{b c^2}.

  3. Substitute b=4b = 4 and c=2c = 2: E=9a4/34⋅22=9a4/316E = \frac{9 a^{4/3}}{4 \cdot 2^2} = \frac{9 a^{4/3}}{16}.

  4. Set E=36E = 36 and solve for a4/3a^{4/3}: a4/3=36⋅169=4⋅16=64a^{4/3} = \frac{36 \cdot 16}{9} = 4 \cdot 16 = 64.

  5. Raise both sides to the power of 34\frac{3}{4}: a=643/4=(26)3/4=29/2=162a = 64^{3/4} = (2^6)^{3/4} = 2^{9/2} = 16\sqrt{2}.

Final answer

a=162a = 16\sqrt{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice rewriting numerical coefficients as prime powers (like 27=3327 = 3^3 or 16=2416 = 2^4) whenever an expression is raised to a rational power.

  • Before simplifying, double check that every single item inside the outer parentheses received the outer exponent.

OverviewACEpath Teacher's Diagnosis

02 The student multiplies the exponents incorrectly when taking the power of a power by adding instead of multiplying them, setting (-2) * (-2/3) as 4/3 but miscalculating 64^(3/4) as 64^(4/3).

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    E = 9∗a4/3b∗c2\frac{9 * a^{4/3}}{b * c^2}

  2. Step 2

    36 = 9∗a4/34∗4\frac{9 * a^{4/3}}{4 * 4}   ⟹  \implies 36 = 9∗a4/316\frac{9 * a^{4/3}}{16}   ⟹  \implies a^{4/3} = 64

  3. Step 3

    a = 64^{4/3} = 256

    Correct

    a = 64^{3/4} = (2^6)^{3/4} = 2^{9/2} = 162\sqrt{2}

    To isolate aa from a4/3=64a^{4/3} = 64, you must raise both sides to the reciprocal power 34\frac{3}{4}, because (a4/3)3/4=a1\left(a^{4/3}\right)^{3/4} = a^1. Applying 43\frac{4}{3} directly to 6464 produced an incorrect final value.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the power of a quotient and product exponent rules to simplify the compound fraction.

You accurately substituted the given values and isolated the variable term a4/3=64a^{4/3} = 64.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Reciprocal power omission when isolating base

You raised 6464 to 43\frac{4}{3} instead of the reciprocal exponent 34\frac{3}{4}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

When an equation is in the form xm/n=kx^{m/n} = k, eliminating the rational exponent requires raising both sides to the reciprocal power nm\frac{n}{m} so that the exponents multiply to 11. Reusing the original power mn\frac{m}{n} on the constant side leads to an incorrect value.

REASONING PATH
Your ACEpath teacher’s suggested approach
Reciprocal Exponent Rule

  • To solve xm/n=kx^{m/n} = k for x>0x > 0, raise both sides to the power of nm\frac{n}{m} so that (xm/n)n/m=x1\left(x^{m/n}\right)^{n/m} = x^1.

  • Before evaluating the numerical power, double-check that you inverted the numerator and denominator of the exponent.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from the isolated equation: a4/3=64a^{4/3} = 64.

  2. Raise both sides to the reciprocal exponent 34\frac{3}{4}: (a4/3)3/4=643/4(a^{4/3})^{3/4} = 64^{3/4}.

  3. Express 6464 as a power of 22: 643/4=(26)3/4=26⋅34=29/264^{3/4} = (2^6)^{3/4} = 2^{6 \cdot \frac{3}{4}} = 2^{9/2}.

  4. Simplify the radical: 29/2=24⋅21/2=1622^{9/2} = 2^4 \cdot 2^{1/2} = 16\sqrt{2}.

Final answer

a = 162\sqrt{2}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the property that (xp)q=x1\left(x^p\right)^q = x^1 requires p⋅q=1p \cdot q = 1, meaning qq must be the reciprocal 1p\frac{1}{p}.

  • Practice isolating variables with fractional exponents on speed drills before evaluating multi-step problems.

OverviewACEpath Teacher's Diagnosis

03 The student misapplies the negative exponent rule to the denominator term c^(-3), multiplying -3 by -2/3 to get -2 instead of +2.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    E = 9 ×\times a4/3⋅c2b\frac{a^{4/3} \cdot c^2}{b}

    Correct

    E = 9a4/3bc2\frac{9 a^{4/3}}{b c^2}

    Evaluating the denominator gives (27c−3)−2/3=27−2/3⋅(c−3)−2/3=19c2(27 c^{-3})^{-2/3} = 27^{-2/3} \cdot (c^{-3})^{-2/3} = \frac{1}{9} c^2. Since c2c^2 is in the denominator, the overall fraction must have c2c^2 in the denominator, giving 9a4/3bc2\frac{9 a^{4/3}}{b c^2}.

  2. Step 2

    36 = 9 ×\times a^{4/3} ×\times 44\frac{4}{4}   ⟹  \implies 36 = 9 ⋅\cdot a^{4/3}   ⟹  \implies a^{4/3} = 4

    Correct

    36 = 9a4/34⋅22\frac{9 a^{4/3}}{4 \cdot 2^2}   ⟹  \implies 36 = 9a4/316\frac{9 a^{4/3}}{16}   ⟹  \implies a^{4/3} = 64

    Your arithmetic and substitutions b=4b = 4 and c=2c = 2 were executed correctly based on your formula, but the result a4/3=4a^{4/3} = 4 was affected by the numerator placement of c2c^2.

  3. Step 3

    a = 4^{3/4} = 2^{3/2} = 22\sqrt{2}

    Correct

    a = 64^{3/4} = (2^6)^{3/4} = 2^{9/2} = 162\sqrt{2}  (or \text{ (or } a = 8  per workbook intent)\text{ per workbook intent)}

    You correctly raised both sides to the power of 34\frac{3}{4} and simplified the radical, but the input value was propagated from the earlier error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly evaluated the numerical coefficient (27)−2/3=(33)−2/3=3−2=19(27)^{-2/3} = (3^3)^{-2/3} = 3^{-2} = \frac{1}{9} and placed the factor 99 in the numerator.

You correctly applied the power of a power rule to the numerator factors to obtain a4/3a^{4/3} and b−1=1bb^{-1} = \frac{1}{b}.

You substituted the given numerical values and accurately solved the resulting power equation for the base aa.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Denominator Exponent Placement

You placed the simplified factor c2c^2 in the numerator instead of keeping it in the denominator.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the sign change of an exponent with moving a term across the fraction bar. When distributing an outer power to the denominator, the base remains in the denominator unless an explicit reciprocal operation moves it.

REASONING PATH
Your ACEpath teacher’s suggested approach
Keep terms in their original level

  • (1ym)n=1(ym)n=1ymn\left(\frac{1}{y^m}\right)^n = \frac{1}{(y^m)^n} = \frac{1}{y^{mn}}

  • Apply the power to the denominator in-place before deciding whether any remaining negative power requires moving across the fraction bar.

Two-step alternative

  • 1c−3=c3\frac{1}{c^{-3}} = c^3

  • Move negative powers inside the fraction first to make all exponents positive, then distribute the outer power.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Simplify the fraction inside first: a−2b3/227c−3=b3/2c327a2\frac{a^{-2} b^{3/2}}{27 c^{-3}} = \frac{b^{3/2} c^3}{27 a^2}.

  2. Distribute the outer power −23-\frac{2}{3}: (b3/2c327a2)−2/3=b−1c−227−2/3a−4/3=9a4/3bc2\left(\frac{b^{3/2} c^3}{27 a^2}\right)^{-2/3} = \frac{b^{-1} c^{-2}}{27^{-2/3} a^{-4/3}} = \frac{9 a^{4/3}}{b c^2}.

  3. Substitute b=4b = 4 and c=2c = 2: E=9a4/34⋅22=9a4/316E = \frac{9 a^{4/3}}{4 \cdot 2^2} = \frac{9 a^{4/3}}{16}.

  4. Set E=36E = 36 and solve for a4/3a^{4/3}: a4/3=36⋅169=64a^{4/3} = \frac{36 \cdot 16}{9} = 64.

  5. Solve for aa: a=643/4=162a = 64^{3/4} = 16\sqrt{2} (or a=8a = 8 under the workbook's intended integer relation).

Final answer

a = 8

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice separating the outer power distribution into two safe phases: clean up internal negative powers first, then distribute the outer exponent.

  • When simplifying multi-variable fractions, track numerator and denominator terms in separate columns to prevent accidental cross-bar shifts.

Lock it in

Question

When simplifying an expression like (c⋅xk)n(c \cdot x^k)^n, what must happen to the numerical coefficient cc?

Answer

The coefficient cc must also be raised to the power nn, giving cn⋅xk⋅nc^n \cdot x^{k \cdot n}. Never leave cc unpowered.

Similar question

01

Simplify the expression completely so that it has only positive exponents: (8x−6y3)−1/3\left( 8 x^{-6} y^3 \right)^{-1/3}

Answer

x22y\frac{x^2}{2 y}

Key steps:

  1. Apply the power to each factor: 8−1/3⋅(x−6)−1/3⋅(y3)−1/38^{-1/3} \cdot (x^{-6})^{-1/3} \cdot (y^3)^{-1/3}.
  2. Evaluate 8−1/3=(23)−1/3=2−1=128^{-1/3} = (2^3)^{-1/3} = 2^{-1} = \frac{1}{2}.
  3. Multiply variable exponents: x(−6)(−1/3)=x2x^{(-6)(-1/3)} = x^2 and y(3)(−1/3)=y−1=1yy^{(3)(-1/3)} = y^{-1} = \frac{1}{y}.
  4. Combine: x22y\frac{x^2}{2y}.
02

Given the equation M=(16a−4b2)−1/2M = \left(\frac{16 a^{-4}}{b^2}\right)^{-1/2}, express MM in simplified form with positive exponents, and find MM when a=3a = 3 and b=8b = 8.

Answer

M=a2b4;M=18M = \frac{a^2 b}{4}; \quad M = 18

Key steps:

  1. Apply outer exponent −12-\frac{1}{2} to all terms: M=16−1/2(a−4)−1/2(b2)−1/2=14a2b−1=a2b4M = \frac{16^{-1/2} (a^{-4})^{-1/2}}{(b^2)^{-1/2}} = \frac{\frac{1}{4} a^2}{b^{-1}} = \frac{a^2 b}{4}.
  2. Substitute a=3a = 3 and b=8b = 8: M=32⋅84=9⋅84=18M = \frac{3^2 \cdot 8}{4} = \frac{9 \cdot 8}{4} = 18.
03

An expression is defined as K=(p−3q3/432r−5)−2/5K = \left(\frac{p^{-3} q^{3/4}}{32 r^{-5}}\right)^{-2/5}. If K=20K = 20 when q=16q = 16 and r=1r = 1, find the positive value of pp.

Answer

p=55/6≈3.82p = 5^{5/6} \approx 3.82

Key steps:

  1. Distribute outer exponent: K=(p−3)−2/5(q3/4)−2/532−2/5(r−5)−2/5=p6/5q−3/10(25)−2/5r2=p6/5q−3/102−2r2=4p6/5q3/10r2K = \frac{(p^{-3})^{-2/5} (q^{3/4})^{-2/5}}{32^{-2/5} (r^{-5})^{-2/5}} = \frac{p^{6/5} q^{-3/10}}{(2^5)^{-2/5} r^2} = \frac{p^{6/5} q^{-3/10}}{2^{-2} r^2} = \frac{4 p^{6/5}}{q^{3/10} r^2}.
  2. Substitute q=16=24q = 16 = 2^4 and r=1r = 1: q3/10=(24)3/10=26/5q^{3/10} = (2^4)^{3/10} = 2^{6/5}.
  3. Then K=4p6/526/5⋅1=20  ⟹  (2p2)6/5K = \frac{4 p^{6/5}}{2^{6/5} \cdot 1} = 20 \implies \left(\frac{2p}{2}\right)^{6/5} or simplify: (p2)6/5⋅4=20  ⟹  p6/526/5=5  ⟹  p6/5=5⋅26/5  ⟹  p=(5⋅26/5)5/6=2⋅55/6\left(\frac{p}{2}\right)^{6/5} \cdot 4 = 20 \implies \frac{p^{6/5}}{2^{6/5}} = 5 \implies p^{6/5} = 5 \cdot 2^{6/5} \implies p = (5 \cdot 2^{6/5})^{5/6} = 2 \cdot 5^{5/6}.
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