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In a biological scaling model, an organism's metabolic rate RR (measured in milliwatts, mW\text{mW}) relates to its body mass MM (measured in grams, g\text{g}) by the equation R=kM3/4R = k M^{3/4}, where k=3.6×10−1 mW⋅g−3/4k = 3.6 \times 10^{-1} \text{ mW} \cdot \text{g}^{-3/4}. Convert the organism's mass to kilograms (M=1.6×10−1 kgM = 1.6 \times 10^{-1} \text{ kg}), express the formula using negative powers of 10 to clear all denominators, and calculate RR in milliwatts to two significant figures.

Answer and steps

Final answer

16 mW16 \text{ mW}

Steps

01

Convert the body mass to grams

The model requires the mass MM to be in grams (g\text{g}), but the value is given in kilograms (kg\text{kg}). Since 1 kg=103 g1\text{ kg} = 10^3\text{ g}:

M=(1.6×10−1 kg)×(103 g⋅kg−1)=1.6×10−1+3 g=1.6×102 g=160 gM = (1.6 \times 10^{-1}\text{ kg}) \times (10^3\text{ g}\cdot\text{kg}^{-1}) = 1.6 \times 10^{-1 + 3}\text{ g} = 1.6 \times 10^2\text{ g} = 160\text{ g}

02

Evaluate M3/4M^{3/4} using rules of exponents

Rewrite 160160 as 16×10116 \times 10^1 so that the rational power can be applied easily to a power of 22 (16=2416 = 2^4):

M3/4=(16×101)3/4=(16)3/4×(101)3/4M^{3/4} = (16 \times 10^1)^{3/4} = (16)^{3/4} \times (10^1)^{3/4}

Apply the power of a power rule, (am)n=am⋅n(a^m)^n = a^{m \cdot n}:

(16)3/4=(24)3/4=24⋅34=23=8(16)^{3/4} = (2^4)^{3/4} = 2^{4 \cdot \frac{3}{4}} = 2^3 = 8

Thus:

M3/4=8×103/4=8×100.75M^{3/4} = 8 \times 10^{3/4} = 8 \times 10^{0.75}

Using the numerical approximation 100.75≈5.623410^{0.75} \approx 5.6234:

M3/4≈8×5.6234≈44.987 g3/4M^{3/4} \approx 8 \times 5.6234 \approx 44.987\text{ g}^{3/4}

03

Calculate the metabolic rate RR

Substitute k=3.6×10−1 mW⋅g−3/4k = 3.6 \times 10^{-1}\text{ mW}\cdot\text{g}^{-3/4} and M3/4M^{3/4} into the equation R=kM3/4R = k M^{3/4}:

R=(3.6×10−1)×(8×100.75)R = (3.6 \times 10^{-1}) \times (8 \times 10^{0.75})

Combine the coefficients and powers of 10 using the product rule 10a⋅10b=10a+b10^a \cdot 10^b = 10^{a+b}:

R=(3.6×8)×10−1+0.75=28.8×10−0.25R = (3.6 \times 8) \times 10^{-1 + 0.75} = 28.8 \times 10^{-0.25}

Evaluating numerically:

10−0.25≈0.5623410^{-0.25} \approx 0.56234

R≈28.8×0.56234≈16.196 mWR \approx 28.8 \times 0.56234 \approx 16.196\text{ mW}

Rounding to two significant figures gives:

R≈16 mWR \approx 16\text{ mW}

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student substitutes the mass directly in kilograms instead of converting to grams to match the unit of k, resulting in an incorrect metabolic rate.

Sample Student Work

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

    M = 1.6 ×\times 10^{-1}   ⟹  \implies R = 3.6 ×\times 10^{-1} ×\times (1.6 ×\times 10^{-1})^{3/4}

    Correct

    M = (1.6 ×\times 10^{-1} kg\text{ kg}) ×\times 10^3 g/kg\text{ g/kg} = 160 g\text{ g}   ⟹  \implies R = (3.6 ×\times 10^{-1}) ×\times (160)^{3/4}

    The constant kk has units of mW⋅g−3/4\text{mW} \cdot \text{g}^{-3/4}, which requires the mass MM to be measured in grams. You substituted the mass directly as 1.6×10−11.6 \times 10^{-1} (the value in kilograms) without multiplying by 103 g/kg10^3\text{ g/kg} to convert it to 160 g160\text{ g}.

  2. Step 2

    (0.16)^{3/4} ≈\approx 0.253   ⟹  \implies R = 0.36 ×\times 0.253 ≈\approx 0.091 mW\text{ mW}

    Correct

    R≈16 mWR \approx 16\text{ mW}

    Your exponent evaluation and final multiplication were executed consistently with your substituted number, but the final value is off by a factor of (103)3/4=109/4≈177.8(10^3)^{3/4} = 10^{9/4} \approx 177.8 due to the missing unit conversion in the first step.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified and applied the scaling formula R=kM3/4R = k M^{3/4}.

You accurately converted scientific notation to decimal form to compute the rational power.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Missing unit conversion before substitution

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the numerical value of mass as unitless rather than checking whether the given units (kilograms) matched the dimensional unit required by the scaling constant kk (grams).

REASONING PATH
Your ACEpath teacher’s suggested approach
Match units to formula constants

  • Every variable substituted into an empirical formula must match the exact base units defined in the constant's unit label.

  • Always write down the units alongside each variable and verify that kg\text{kg} is converted to g\text{g} whenever a constant contains g−n\text{g}^{-n}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Convert the mass from kilograms to grams: M=1.6×10−1 kg×103 g/kg=1.6×102 g=160 gM = 1.6 \times 10^{-1}\text{ kg} \times 10^3\text{ g/kg} = 1.6 \times 10^2\text{ g} = 160\text{ g}.

  2. Rewrite 160 g160\text{ g} as 16×10116 \times 10^1 and apply the rational power: (160)3/4=(16)3/4×103/4=8×100.75≈44.987 g3/4(160)^{3/4} = (16)^{3/4} \times 10^{3/4} = 8 \times 10^{0.75} \approx 44.987\text{ g}^{3/4}.

  3. Multiply by k=3.6×10−1 mW⋅g−3/4k = 3.6 \times 10^{-1}\text{ mW}\cdot\text{g}^{-3/4}: R=(3.6×10−1)×44.987≈16.196 mWR = (3.6 \times 10^{-1}) \times 44.987 \approx 16.196\text{ mW}.

  4. Round to two significant figures: R≈16 mWR \approx 16\text{ mW}.

Final answer

16 mW16\text{ mW}

NEXT STEPS
Final Takeaways from Your Teacher

  • Before plugging values into any power law or scaling model, underline the units of each constant to confirm what unit each variable must have.

  • Practice using exponent properties on powers of 10 to convert metric prefixes systematically.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly divides the mass by 1000 instead of multiplying when converting from kilograms to grams, obtaining M = 1.6 × 10⁻⁴ g.

Sample Student Work

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

    M = 1.6 ×\times 10^{-1} ×\times 10^{-3} g\text{ g} = 1.6 ×\times 10^{-4} g\text{ g}

    Correct

    M = 1.6 ×\times 10^{-1} kg\text{ kg} ×\times 10^3\text{ g}$$\cdot$$\text{kg}^{-1} = 1.6 ×\times 10^2 g\text{ g} = 160 g\text{ g}

    To convert from kilograms to grams, you need to multiply by 103 g/kg10^3\text{ g}/\text{kg} rather than 10−310^{-3}, because one kilogram contains one thousand grams.

  2. Step 2

    R = 3.6 ×\times 10^{-1} ×\times (1.6 ×\times 10^{-4})^{3/4}

    Correct

    R = (3.6 ×\times 10^{-1}) ×\times (160)^{3/4}

    You correctly substituted your calculated mass into the model equation R=kM3/4R = k M^{3/4}, but the mass value was carried over from the incorrect conversion in Step 1.

  3. Step 3

    (1.6 ×\times 10^{-4})^{3/4} ≈\approx 0.00253   ⟹  \implies R = 3.6 ×\times 10^{-1} ×\times 0.00253 ≈\approx 0.00091 mW\text{ mW}

    Correct

    R ≈\approx 3.6 ×\times 10^{-1} ×\times 44.987 ≈\approx 16 mW\text{ mW}

    Your numerical evaluation of the rational power and multiplication were calculated consistently, but the result is off due to the initial mass conversion error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified and substituted the parameters into the allometric equation R=kM3/4R = k M^{3/4}.

You accurately carried out the fractional power evaluation (M)3/4(M)^{3/4} with your substituted value.

You applied proper scientific rounding to two significant figures.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Inverted metric unit conversion factor

You multiplied by 10−310^{-3} instead of 10310^3 when converting kilograms to grams.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the direction of metric prefix conversion: applying 10−310^{-3} (the conversion from grams to kilograms) when converting from a larger unit (kilograms) to a smaller unit (grams).

REASONING PATH
Your ACEpath teacher’s suggested approach
Unit Size Check

  • Converting from a larger unit to a smaller unit must yield a numerically larger quantity (1 kg=103 g1\text{ kg} = 10^3\text{ g}).

  • Always multiply by 103 g/kg10^3\text{ g}/\text{kg} when going from kg\text{kg} to g\text{g}, making the exponent increase by +3+3.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Convert mass correctly: M=1.6×10−1 kg×103 g/kg=1.6×102 g=160 gM = 1.6 \times 10^{-1}\text{ kg} \times 10^3\text{ g}/\text{kg} = 1.6 \times 10^2\text{ g} = 160\text{ g}.

  2. Evaluate the rational power: M3/4=(16×101)3/4=163/4×100.75=8×100.75≈44.987 g3/4M^{3/4} = (16 \times 10^1)^{3/4} = 16^{3/4} \times 10^{0.75} = 8 \times 10^{0.75} \approx 44.987\text{ g}^{3/4}.

  3. Multiply by kk: R=(3.6×10−1)×44.987=16.195... mWR = (3.6 \times 10^{-1}) \times 44.987 = 16.195...\text{ mW}.

Final answer

16 mW16\text{ mW}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice metric prefix conversions by explicitly writing out conversion fractions with canceling units.

  • Check whether your converted numerical value should be larger or smaller before proceeding with calculations.

OverviewACEpath Teacher's Diagnosis

03 The student confuses the fractional exponent 3/4 with the multiplier 3/4, multiplying M by 3/4 instead of raising it to the power of 3/4.

Sample Student Work

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

    M=1.6×10−1×103 g=160 gM = 1.6 \times 10^{-1} \times 10^3\text{ g} = 160\text{ g}

  2. Step 2

    R=3.6×10−1×(34×160)  ⟹  34×160=120R = 3.6 \times 10^{-1} \times \left(\frac{3}{4} \times 160\right) \implies \frac{3}{4} \times 160 = 120

    Correct

    M3/4=1603/4=(16×10)3/4=163/4×100.75=8×100.75≈45.0M^{3/4} = 160^{3/4} = (16 \times 10)^{3/4} = 16^{3/4} \times 10^{0.75} = 8 \times 10^{0.75} \approx 45.0

    You multiplied the base 160160 by the fraction 34\frac{3}{4} instead of raising 160160 to the power of 34\frac{3}{4}.

  3. Step 3

    R=0.36×120=43 mWR = 0.36 \times 120 = 43\text{ mW}

    Correct

    R=3.6×10−1×1603/4≈16 mWR = 3.6 \times 10^{-1} \times 160^{3/4} \approx 16\text{ mW}

    Your arithmetic 0.36×120=43.2≈43 mW0.36 \times 120 = 43.2 \approx 43\text{ mW} was carried out consistently, but it inherited the error from treating the exponent as a coefficient.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly converted the organism's mass from kilograms to grams using powers of 10.

You accurately substituted your computed intermediate value into the linear multiplier k=0.36k = 0.36.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Exponent treated as scalar multiplier

You wrote (34×160)(\frac{3}{4} \times 160) where the model required 1603/4160^{3/4}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You confused rational exponentiation xp/q=xpqx^{p/q} = \sqrt[q]{x^p} with scalar multiplication pq⋅x\frac{p}{q} \cdot x.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distinguish Exponent vs. Coefficient

  • bm/n≠mn⋅bb^{m/n} \neq \frac{m}{n} \cdot b; rather, bm/n=(bn)mb^{m/n} = (\sqrt[n]{b})^m

  • Always keep the rational fraction in the superscript position and evaluate it as a root followed by an integer power.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Rewrite the mass M=160 gM = 160\text{ g} as 16×101 g16 \times 10^1\text{ g}.

  2. Apply the exponent rule to each factor: M3/4=163/4×103/4=(24)3/4×100.75=23×100.75=8×100.75M^{3/4} = 16^{3/4} \times 10^{3/4} = (2^4)^{3/4} \times 10^{0.75} = 2^3 \times 10^{0.75} = 8 \times 10^{0.75}.

  3. Multiply by k=3.6×10−1k = 3.6 \times 10^{-1}: R=(3.6×10−1)×(8×100.75)=28.8×10−0.25R = (3.6 \times 10^{-1}) \times (8 \times 10^{0.75}) = 28.8 \times 10^{-0.25}.

  4. Evaluate numerically: 28.8×0.56234≈16.196 mW28.8 \times 0.56234 \approx 16.196\text{ mW}, which rounds to 16 mW16\text{ mW}.

Final answer

16 mW16\text{ mW}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice separating a base into clean powers when working with fractional exponents (for example, 1603/4=(16×10)3/4160^{3/4} = (16 \times 10)^{3/4}).

  • Double-check that fractional numbers written in exponents never drop down into the baseline as multiplying fractions.

Lock it in

Question

Why must you check the unit of a proportionality constant kk before substituting variables into a power-law equation like y=kxpy = k x^p?

Answer

The unit of kk dictates the specific units required for xx and yy so that all units balance properly when evaluated.

Similar question

01

An allometric relation gives surface area S=4.0×M2/3S = 4.0 \times M^{2/3}, where SS is in cm2\text{cm}^2 and MM is in grams (g\text{g}). If an organism has mass M=8.0×10−2 kgM = 8.0 \times 10^{-2}\text{ kg}, calculate SS in cm2\text{cm}^2.

Answer

S≈74 cm2S \approx 74\text{ cm}^2. First convert mass to grams: M=8.0×10−2 kg×103 g/kg=80 gM = 8.0 \times 10^{-2}\text{ kg} \times 10^3\text{ g/kg} = 80\text{ g}. Then evaluate S=4.0×(80)2/3=4.0×18.566≈74.26 cm2S = 4.0 \times (80)^{2/3} = 4.0 \times 18.566 \approx 74.26\text{ cm}^2.

02

A scaling constant is given as C=2.5×10−2 J⋅m−3/2C = 2.5 \times 10^{-2}\text{ J}\cdot\text{m}^{-3/2}. A biological specimen has length L=4.0 cmL = 4.0\text{ cm}. Express LL in meters using negative powers of 10, then find the energy E=CL3/2E = C L^{3/2} in scientific notation.

Answer

E=2.0×10−4 JE = 2.0 \times 10^{-4}\text{ J}. Convert LL to meters: L=4.0×10−2 mL = 4.0 \times 10^{-2}\text{ m}. Compute L3/2=(4.0×10−2)3/2=(4.0)3/2×(10−2)3/2=8.0×10−3 m3/2L^{3/2} = (4.0 \times 10^{-2})^{3/2} = (4.0)^{3/2} \times (10^{-2})^{3/2} = 8.0 \times 10^{-3}\text{ m}^{3/2}. Finally, E=(2.5×10−2)×(8.0×10−3)=20.0×10−5=2.0×10−4 JE = (2.5 \times 10^{-2}) \times (8.0 \times 10^{-3}) = 20.0 \times 10^{-5} = 2.0 \times 10^{-4}\text{ J}.

03

The metabolic equation R=kM3/4R = k M^{3/4} uses k=0.36 mW⋅g−3/4k = 0.36\text{ mW}\cdot\text{g}^{-3/4}. A researcher erroneously plugs in M=0.50 kgM = 0.50\text{ kg} directly without unit conversion. By what numerical factor is their calculated RR smaller than the true value?

Answer

The calculated RR is smaller by a factor of 109/4=102.25≈177.810^{9/4} = 10^{2.25} \approx 177.8. Since 1 kg=103 g1\text{ kg} = 10^3\text{ g}, omitting the factor of 10310^3 inside the power (M)3/4(M)^{3/4} underestimates the result by (103)3/4=109/4≈178(10^3)^{3/4} = 10^{9/4} \approx 178.

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