In an analytical chemistry laboratory, a spectrophotometer's transmittance $T$ is related to absorbance $A$ by $T = 10^{-A}$. An automated sensor measures the output voltage $V$ (in millivolts) governed by the empirical power relationship $V = 250 \cdot T^{3/4}$. During the standardization of an unknown dye solution, the optical absorbance is determined to be $A = 0.80$. Using the properties of negative and rational exponents, calculate the detector output voltage $V$ in millivolts to two significant figures.

In an analytical chemistry laboratory, a spectrophotometer's transmittance $T$ is related to absorbance $A$ by $T = 10^{-A}$ . An automated sensor measures the output voltage $V$ (in millivolts) governed by the empirical power relationship $V = 250 \cdot T^{3/4}$ . During the standardization of an unknown dye solution, the optical absorbance is determined to be $A = 0.80$ . Using the properties of negative and rational exponents, calculate the detector output voltage $V$ in millivolts to two significant figures.

Final answer

$63 \text{ mV}$

Steps

Combine the equations using the power of a power rule

To find the detector output voltage $V$ , substitute the transmittance relationship into the voltage equation and simplify the exponents.

The output voltage is given by: $V = 250 \cdot T^{3/4}$

Since $T = 10^{-A}$ , substitute $10^{-A}$ for $T$ : $V = 250 \cdot (10^{-A})^{3/4}$

Applying the power of a power property, $(b^m)^n = b^{m \cdot n}$ : $V = 250 \cdot 10^{-A \cdot \frac{3}{4}}$

Substitute the absorbance value and simplify the exponent

Substitute the measured absorbance $A = 0.80 = \frac{4}{5}$ : $-A \cdot \frac{3}{4} = -\frac{4}{5} \cdot \frac{3}{4} = -\frac{3}{5} = -0.60$

Thus, the formula simplifies to: $V = 250 \cdot 10^{-0.60}$

Using the negative exponent property $b^{-n} = \frac{1}{b^n}$ : $V = \frac{250}{10^{0.60}} = \frac{250}{10^{3/5}}$

Tip: Simplifying the exponents symbolically before calculating numerical values prevents intermediate rounding errors and cancels out the factor of $4$ directly: $\left(\frac{4}{5}\right) \cdot \left(\frac{3}{4}\right) = \frac{3}{5}$ .

Compute the numerical value and round

Calculate the power of ten: $10^{-0.60} \approx 0.25119$

Multiply by $250$ : $V = 250 \cdot 0.25119 \approx 62.80\text{ mV}$

Rounding to two significant figures gives: $V \approx 63\text{ mV}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student multiplies the base by the exponent instead of evaluating the power, computing 10 · (-0.60) = -6.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = 10^{(-0.80)} $\implies$ V = 250 $\cdot$ (10^{(-0.80)})^{ $\frac{3}{4}$ } $\implies$ V = 250 $\cdot$ 10^{(-0.60)}
Step 2
10^{(-0.60)} = 10 $\cdot$ (-0.60) = -6.0
Correct
10^{-0.60} = $\frac{1}{10^{0.60}}$ $\approx$ 0.2512
An exponent represents repeated multiplication or power scaling, not multiplication of the base by the exponent. A positive base raised to any real power always yields a positive result.
Step 3
V = 250 $\cdot$ (-6.0) = -1500 $\text{ mV}$
Correct
V = 250 $\cdot$ 10^{-0.60} $\approx$ 250 $\cdot$ 0.2512 $\approx$ 63 $\text{ mV}$
This calculation correctly multiplied $250$ by your previous step's result, but it inherited the incorrect value of $-6.0$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You substituted the absorbance value $A = 0.80$ into the transmittance relationship correctly.

You accurately combined powers using the power of a power property $(b^m)^n = b^{m \cdot n}$ to get $10^{-0.60}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Base-exponent multiplication trap

You replaced an exponential expression $b^x$ with multiplication $b \cdot x$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing exponentiation with multiplication when dealing with negative and decimal powers, overlooking that $b^x > 0$ for any positive base $b$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent Definition Check: The expression $b^n$ indicates the base $b$ raised to the power $n$ , never $b \cdot n$ .
Rewrite negative powers using reciprocal form $b^{-n} = \frac{1}{b^n}$ before evaluating.
Sign Feasibility Check: For any positive base $b > 0$ and real number $x$ , $b^x$ is strictly positive ( $b^x > 0$ ).
If your power evaluation produces a negative number from a positive base, pause and recalculate.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite $10^{-0.60}$ as $\frac{1}{10^{0.60}}$ using the negative exponent rule $b^{-n} = \frac{1}{b^n}$ .
Evaluate the power: $10^{-0.60} \approx 0.25119$ .
Multiply by the constant: $V = 250 \cdot 0.25119 \approx 62.80\text{ mV}$ .
Round to two significant figures to obtain $63\text{ mV}$ .

Final answer

$63\text{ mV}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice converting negative exponents into fractions before computing their decimal values.
Verify that powers of positive numbers always yield positive values before carrying out final multiplications.

OverviewACEpath Teacher's Diagnosis

02 The student ignores the negative sign in the exponent and computes 10^(0.60) instead of 10^(-0.60), resulting in an inflated voltage of 990 mV.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = 10^{-0.80} $\implies$ V = 250 $\cdot$ (10^{-0.80})^{3/4}
Step 2
V = 250 $\cdot$ 10^{0.60}
Correct
V = 250 $\cdot$ 10^{-0.60}
When applying the power of a power rule $(b^m)^n = b^{m \cdot n}$ , multiplying a negative exponent by a positive exponent yields a negative result: $(-0.80) \cdot \frac{3}{4} = -0.60$ , not $+0.60$ .
Step 3
10^{0.60} $\approx$ 3.98 $\implies$ V = 250 $\cdot$ 3.98 = 995 $\approx$ 990 $\text{ mV}$
Correct
V = 250 $\cdot$ 10^{-0.60} $\approx$ 250 $\cdot$ 0.2512 = 62.8 $\approx$ 63 $\text{ mV}$
Because the exponent was missing its negative sign, the power evaluated to approximately $3.98$ rather than $0.2512$ , propagating to an incorrect final voltage.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the absorbance into the transmittance formula.

You properly structured the detector output voltage equation using the power of a power rule.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error in exponent multiplication

You dropped the negative sign when multiplying $-0.80$ by $\frac{3}{4}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the power multiplication as magnitude-only arithmetic, overlooking that a negative base exponent must preserve its negative sign under a positive fractional power.

REASONING PATH

Your ACEpath teacher’s suggested approach

Sign tracking in power of a power: $(b^{-m})^n = b^{-(m \cdot n)}$ when $n > 0$ .
Check the sign of each factor before computing the numerical product in the exponent.
Physical magnitude check: Transmittance is between $0$ and $1$ for positive absorbance, so $T^{3/4} < 1$ .
Ensure that the factor multiplying $250$ reduces the voltage rather than scaling it up.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply the exponents while keeping the negative sign: $(-0.80) \cdot \frac{3}{4} = -0.60$ .
Rewrite with a negative exponent: $V = 250 \cdot 10^{-0.60}$ .
Evaluate the negative power of ten: $10^{-0.60} = \frac{1}{10^{0.60}} \approx 0.25119$ .
Multiply by the coefficient: $V \approx 250 \cdot 0.25119 = 62.797 \text{ mV}$ .

Final answer

$63 \text{ mV}$

NEXT STEPS

Final Takeaways from Your Teacher

Write out signed products explicitly, such as $(-a) \cdot b = -(ab)$ , before evaluating powers of ten.
Perform a sanity check on exponential decay problems: positive absorbance means transmittance is less than $1$ , so the output voltage must be less than $250\text{ mV}$ .

OverviewACEpath Teacher's Diagnosis

03 The student adds the exponents instead of multiplying them when applying the power of a power rule, calculating -0.80 + 0.75 = -0.05.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = 10^{-0.80} $\implies$ V = 250 $\cdot$ (10^{-0.80})^{3/4}
Step 2
V = 250 $\cdot$ 10^{-0.80 + 0.75} $\implies$ V = 250 $\cdot$ 10^{-0.05}
Correct
V = 250 $\cdot$ 10^{-0.80 $\cdot$ 0.75} = 250 $\cdot$ 10^{-0.60}
When raising a power to another power, you must multiply the exponents according to $(b^m)^n = b^{m \cdot n}$ . Here, the exponent should be $-0.80 \times 0.75 = -0.60$ , not $-0.80 + 0.75 = -0.05$ .
Step 3
10^{-0.05} $\approx$ 0.891 $\implies$ V = 250 $\cdot$ 0.891 = 222.75 $\approx$ 220 $\text{ mV}$
Correct
V = 250 $\cdot$ 10^{-0.60} $\approx$ 250 $\cdot$ 0.25119 $\approx$ 63 $\text{ mV}$
Your arithmetic and rounding follow directly from $10^{-0.05}$ , but the resulting value is incorrect due to the exponent error in the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the given absorbance into the transmittance equation.

You accurately substituted the expression for transmittance into the sensor voltage formula.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Power of a Power vs. Product Rule Confusion

You added $-0.80$ and $0.75$ instead of multiplying them when simplifying $(10^{-0.80})^{3/4}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the power of a power property $(b^m)^n = b^{m \cdot n}$ with the product property $b^m \cdot b^n = b^{m+n}$ . Adding exponents applies only when multiplying terms that share the same base, whereas raising an exponential expression to an outer power requires multiplying the exponents.

REASONING PATH

Your ACEpath teacher’s suggested approach

Rule Check: One Base vs. Two Bases: For a single base raised to nested powers, use $(b^m)^n = b^{m \cdot n}$ . Use $b^m \cdot b^n = b^{m+n}$ only when multiplying two powers with the same base.
Whenever an exponent is outside parentheses, such as $(b^m)^n$ , multiply the inner and outer exponents.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the power of a power rule: $(10^{-0.80})^{3/4} = 10^{(-0.80) \cdot (3/4)} = 10^{-0.60}$ .
Rewrite in fraction form: $-0.60 = -\frac{3}{5}$ , so $V = 250 \cdot 10^{-3/5} = \frac{250}{10^{3/5}}$ .
Compute the numerical power: $10^{-0.60} \approx 0.25119$ .
Multiply by $250$ : $V = 250 \cdot 0.25119 \approx 62.798\text{ mV}$ .

Final answer

$63\text{ mV}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between $(x^a)^b = x^{a \cdot b}$ and $x^a \cdot x^b = x^{a+b}$ using simple integer examples like $(2^2)^3$ versus $2^2 \cdot 2^3$ .
Practice converting decimal exponents into simple fractions (e.g., $0.80 = \frac{4}{5}$ and $0.75 = \frac{3}{4}$ ) to cancel common factors before multiplying.

Lock it in

Question

What does a negative exponent $b^{-n}$ represent mathematically?

Answer

$b^{-n} = \frac{1}{b^n}$ . A negative exponent denotes the reciprocal of the base raised to the positive power, not a negative product.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student multiplies the base by the exponent instead of evaluating the power, computing 10 · (-0.60) = -6.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student ignores the negative sign in the exponent and computes 10^(0.60) instead of 10^(-0.60), resulting in an inflated voltage of 990 mV.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student adds the exponents instead of multiplying them when applying the power of a power rule, calculating -0.80 + 0.75 = -0.05.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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