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In a spectrophotometry lab, absorbance data is modeled by the logarithmic equation A=−log⁡10(II0)A = -\log_{10}\left(\frac{I}{I_0}\right). Solve this equation for the transmitted light intensity II and rewrite the expression in an equivalent base-ee exponential form I=I0⋅e−kAI = I_0 \cdot e^{-k A}, determining the exact constant kk in terms of the natural logarithm of 1010.

Answer and steps

Final answer

I=I0⋅e−(ln⁡(10))AI = I_0 \cdot e^{-(\ln(10))A}, where k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303

Steps

01

Isolate the logarithm and solve for II in base 10

Start with the given absorbance equation (Beer–Lambert law relationship): A=−log⁡10(II0)A = -\log_{10}\left(\frac{I}{I_0}\right)

Multiply both sides by −1-1 to isolate the logarithmic expression: log⁡10(II0)=−A\log_{10}\left(\frac{I}{I_0}\right) = -A

Apply the definition of a logarithm (log⁡b(y)=x  ⟺  y=bx\log_b(y) = x \iff y = b^x): II0=10−A\frac{I}{I_0} = 10^{-A}

Multiply both sides by the incident light intensity I0I_0: I=I0⋅10−AI = I_0 \cdot 10^{-A}

02

Convert the base from 1010 to ee

To rewrite the expression in terms of base ee, use the exponential-logarithm identity for any positive real number bb: b=eln⁡(b)b = e^{\ln(b)}

Setting b=10b = 10: 10=eln⁡(10)10 = e^{\ln(10)}

Substitute this expression into the equation for II: I=I0⋅(eln⁡(10))−AI = I_0 \cdot \left(e^{\ln(10)}\right)^{-A}

Using the power of a power property, (am)n=am⋅n\left(a^m\right)^n = a^{m \cdot n}: I=I0⋅e−(ln⁡(10))AI = I_0 \cdot e^{-(\ln(10))A}

03

Identify the constant kk

Compare this result with the target form I=I0⋅e−kAI = I_0 \cdot e^{-kA}: k=ln⁡(10)k = \ln(10)

Evaluating the constant numerically: k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303

Therefore, the equivalent base-ee expression is: I=I0⋅e−(ln⁡(10))A,where k=ln⁡(10)≈2.303I = I_0 \cdot e^{-(\ln(10))A}, \quad \text{where } k = \ln(10) \approx 2.303

Common Pitfall: When converting bases using bx=exln⁡(b)b^x = e^{x \ln(b)}, make sure not to invert the logarithm as 1ln⁡(10)\frac{1}{\ln(10)} or confuse natural log (ln⁡\ln) with common log (log⁡10\log_{10}). Since 10>e10 > e, the exponent scaling constant k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303 is greater than 11.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets the negative sign when converting from the logarithmic form to the exponential form, resulting in k = -ln 10.

Sample Student Work

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

    A = -\log_{10}(I / I_0)   ⟹  \implies I / I_0 = 10^A   ⟹  \implies I = I_0 ⋅\cdot 10^A_

    Correct

    A = -\log_{10}(I / I_0)   ⟹  \implies \log_{10}(I / I_0) = -A   ⟹  \implies I / I_0 = 10^{-A}   ⟹  \implies I = I_0 ⋅\cdot 10^{-A}

    When isolating the logarithm, multiplying both sides by −1-1 gives log⁡10(I/I0)=−A\log_{10}(I / I_0) = -A. Converting to exponential form therefore yields I/I0=10−AI / I_0 = 10^{-A}, not 10A10^A.

  2. Step 2

    10 = e^{(\ln 10)}   ⟹  \implies I = I_0 ⋅\cdot e^{(A ⋅\cdot \ln 10)}

    Correct

    10 = e^{\ln(10)}   ⟹  \implies I = I_0 ⋅\cdot (e^{\ln(10)})^{-A} = I_0 ⋅\cdot e^{-(\ln(10))A}

    Your substitution 10=eln⁡(10)10 = e^{\ln(10)} is mathematically valid, but the exponent carried over the missing negative sign from Step 1.

  3. Step 3

    I = I_0 ⋅\cdot e^{-(-\ln 10) ⋅\cdot A}   ⟹  \implies k = -\ln 10

    Correct

    I = I_0 ⋅\cdot e^{-kA}   ⟹  \implies k = \ln(10)

    Because of the earlier sign error, you had to rewrite the positive exponent as −(−kA)-(-k A), leading to an extraneous negative sign in k=−ln⁡(10)k = -\ln(10).

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly stated and applied the base-conversion identity 10=eln⁡(10)10 = e^{\ln(10)}.

You skillfully matched the exponent to the required target template I=I0⋅e−kAI = I_0 \cdot e^{-kA}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign Error in Logarithmic Inversion

Dropping the negative sign when converting A=−log⁡10(y)A = -\log_{10}(y) into exponential form.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You exponentiated before fully isolating the logarithm log⁡10(I/I0)\log_{10}(I/I_0), treating −log⁡10(x)-\log_{10}(x) as if it converted directly to 10A10^A rather than moving the negative sign to the other side first.

REASONING PATH
Your ACEpath teacher’s suggested approach
Isolate Before Inverting

  • Before converting log⁡b(M)=N\log_b(M) = N to M=bNM = b^N, ensure the logarithm has a coefficient of +1+1.

  • Always rewrite −y=log⁡b(x)-y = \log_b(x) as log⁡b(x)=−y\log_b(x) = -y before raising the base to both sides.

Physical Reality Check

  • Absorbance reduces transmitted intensity: as absorbance AA increases, II must decrease.

  • Check whether the decay exponent is negative when k>0k > 0, ensuring I<I0I < I_0 for positive absorbance.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Isolate the logarithm by multiplying both sides by −1-1: log⁡10(II0)=−A\log_{10}\left(\frac{I}{I_0}\right) = -A.

  2. Rewrite in exponential form: II0=10−A\frac{I}{I_0} = 10^{-A}, so I=I0⋅10−AI = I_0 \cdot 10^{-A}.

  3. Substitute 10=eln⁡(10)10 = e^{\ln(10)}: I=I0⋅(eln⁡(10))−A=I0⋅e−(ln⁡(10))AI = I_0 \cdot \left(e^{\ln(10)}\right)^{-A} = I_0 \cdot e^{-(\ln(10))A}.

  4. Compare with I=I0⋅e−kAI = I_0 \cdot e^{-kA} to read off kk directly: k=ln⁡(10)k = \ln(10).

Final answer

I=I0⋅e−(ln⁡(10))AI = I_0 \cdot e^{-(\ln(10))A}, where k=ln⁡(10)k = \ln(10)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice isolating logarithmic terms with negative coefficients before converting to exponential form.

  • Verify your final equation against physical boundary conditions, checking that light intensity decreases as absorbance increases.

OverviewACEpath Teacher's Diagnosis

02 The student misapplies the change-of-base relation by writing 10 = e^(1 / ln 10), leading to k = 1 / ln 10.

Sample Student Work

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

    −A=log⁡10(I/I0)  ⟹  II0=10−(−A)  ⟹  I=I0⋅10−(−A)-A = \log_{10}(I/I_0) \implies \frac{I}{I_0} = 10^{-(-A)} \implies I = I_0 \cdot 10^{-(-A)}

    Correct

    −A=log⁡10(II0)  ⟹  II0=10−A  ⟹  I=I0⋅10−A-A = \log_{10}\left(\frac{I}{I_0}\right) \implies \frac{I}{I_0} = 10^{-A} \implies I = I_0 \cdot 10^{-A}

    You correctly multiplied by −1-1 to isolate log⁡10(I/I0)=−A\log_{10}(I/I_0) = -A. However, when converting to exponential form, you introduced an unnecessary extra negative sign, writing 10−(−A)10^{-(-A)}. By the definition of logarithms, log⁡10(y)=x  ⟺  y=10x\log_{10}(y) = x \iff y = 10^x, so with x=−Ax = -A, the correct expression is II0=10−A\frac{I}{I_0} = 10^{-A}.

  2. Step 2

    10=e1ln⁡1010 = e^{\frac{1}{\ln 10}}

    Correct

    10=eln⁡(10)10 = e^{\ln(10)}

    You inverted the natural logarithm in the exponent. Because the exponential function and the natural logarithm are inverse operations, eln⁡(x)=xe^{\ln(x)} = x for any positive xx. Therefore, 10=eln⁡(10)10 = e^{\ln(10)}, not e1ln⁡(10)e^{\frac{1}{\ln(10)}}.

  3. Step 3

    I=I0⋅(e1ln⁡10)−A  ⟹  I=I0⋅e−(1ln⁡10)AI = I_0 \cdot \left(e^{\frac{1}{\ln 10}}\right)^{-A} \implies I = I_0 \cdot e^{-\left(\frac{1}{\ln 10}\right)A}

    Correct

    I=I0⋅(eln⁡(10))−A  ⟹  I=I0⋅e−(ln⁡(10))AI = I_0 \cdot \left(e^{\ln(10)}\right)^{-A} \implies I = I_0 \cdot e^{-(\ln(10))A}

    You correctly applied the power of a power exponent rule (am)n=am⋅n(a^m)^n = a^{m \cdot n} (and reverted to the correct exponent −A-A), but the expression inherits the erroneous base identity from Step 2.

  4. Step 4

    k=1ln⁡10k = \frac{1}{\ln 10}

    Correct

    k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303

    You correctly matched the form I=I0⋅e−kAI = I_0 \cdot e^{-kA} to read off kk, but your final value is the reciprocal of the correct constant due to the error in Step 2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly isolated the logarithmic expression by writing −A=log⁡10(I/I0)-A = \log_{10}(I/I_0).

You recognized that converting from base 10 to base ee requires expressing 1010 as a power of ee.

You applied the power of a power property (am)n=amn(a^m)^n = a^{mn} correctly when multiplying exponents.

You accurately matched your simplified expression with the target template I=I0⋅e−kAI = I_0 \cdot e^{-kA} to isolate kk.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Exponential Base Conversion Identity

Writing 10=e1ln⁡(10)10 = e^{\frac{1}{\ln(10)}} instead of 10=eln⁡(10)10 = e^{\ln(10)}, placing the natural logarithm in the denominator of the exponent.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You confused the fundamental inverse identity b=eln⁡(b)b = e^{\ln(b)} with the change-of-base reciprocal property of logarithms log⁡b(a)=1log⁡a(b)\log_b(a) = \frac{1}{\log_a(b)}. Since exe^x and ln⁡(x)\ln(x) are direct inverses, eln⁡(b)e^{\ln(b)} equals bb directly without inverting the exponent.

REASONING PATH
Your ACEpath teacher’s suggested approach
Direct Inverse Identity

  • For any positive base bb, b=eln⁡(b)b = e^{\ln(b)} because eln⁡(x)=xe^{\ln(x)} = x.

  • To change a base to ee, place ln⁡(b)\ln(b) directly in the numerator of the exponent: bx=(eln⁡(b))x=exln⁡(b)b^x = (e^{\ln(b)})^x = e^{x\ln(b)}.

Magnitude Reality Check

  • Since 10>e≈2.71810 > e \approx 2.718, the exponent must satisfy ln⁡(10)>1\ln(10) > 1.

  • Check whether your scaling constant is larger than 11: ln⁡(10)≈2.303>1\ln(10) \approx 2.303 > 1, whereas 1ln⁡(10)≈0.434<1\frac{1}{\ln(10)} \approx 0.434 < 1, which would give e0.434<e1<10e^{0.434} < e^1 < 10.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the identity b=eln⁡(b)b = e^{\ln(b)} to write 10=eln⁡(10)10 = e^{\ln(10)}.

  2. Substitute this into I=I0⋅10−AI = I_0 \cdot 10^{-A} to get I=I0⋅(eln⁡(10))−AI = I_0 \cdot \left(e^{\ln(10)}\right)^{-A}.

  3. Multiply exponents using (am)n=amn(a^m)^n = a^{mn}: I=I0⋅e−(ln⁡(10))AI = I_0 \cdot e^{-(\ln(10))A}.

  4. Equate with I=I0⋅e−kAI = I_0 \cdot e^{-kA} to obtain k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303.

Final answer

I=I0⋅e−(ln⁡(10))AI = I_0 \cdot e^{-(\ln(10))A}, where k=ln⁡(10)≈2.303k = \ln(10) \approx 2.303

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice converting constant bases (2,5,102, 5, 10) into powers of ee using b=eln⁡(b)b = e^{\ln(b)} until the direct inverse relationship becomes automatic.

  • Use a quick numerical sanity check whenever converting bases: if the original base is greater than ee, the natural log exponent must be greater than 11.

OverviewACEpath Teacher's Diagnosis

03 The student uses natural logarithm directly as the inverse of base-10 logarithm, conflating log10 with ln.

Sample Student Work

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

    −A=log⁡10(II0)-A = \log_{10}\left(\frac{I}{I_0}\right)

  2. Step 2

    II0=e−A\frac{I}{I_0} = e^{-A}

    Correct

    II0=10−A\frac{I}{I_0} = 10^{-A}

    The inverse operation of the common logarithm log⁡10(x)\log_{10}(x) is raising 1010 to that power, not ee. Exponentiating base 1010 gives II0=10−A\frac{I}{I_0} = 10^{-A}.

  3. Step 3

    I = I_0 * e^{(-A)}   ⟹  \implies I = I_0 * e^{(-1 * A)}

    Correct

    I = I_0 ⋅\cdot 10^{-A} = I_0 ⋅\cdot (eln⁡(10))\left(e^{\ln(10)}\right)^{-A} = I_0 ⋅\cdot e^{-(\ln(10))A}

    Your multiplication by I0I_0 followed logically from your previous expression, but it inherited the incorrect base ee.

  4. Step 4

    k = 1

    Correct

    k = \ln(10)

    Because the base was changed directly to ee without the base-conversion factor ln⁡(10)\ln(10), the resulting exponent coefficient became 11 instead of ln⁡(10)\ln(10).

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly isolated the logarithmic term by multiplying through by −1-1.

You correctly recognized that the target form required isolating II and matching the coefficient of −A-A in the exponent.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Conflating Logarithmic Bases

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You directly inverted log⁡10\log_{10} with base ee rather than base 1010, overlooking that log⁡10(y)=x  ⟺  y=10x\log_{10}(y) = x \iff y = 10^x, and that converting 10−A10^{-A} to base ee requires the identity 10=eln⁡(10)10 = e^{\ln(10)}.

REASONING PATH
Your ACEpath teacher’s suggested approach
Matching Logarithm Base to Exponential Base

  • log⁡b(y)=x  ⟺  y=bx\log_b(y) = x \iff y = b^x. The natural base ee only inverts ln⁡(y)=log⁡e(y)\ln(y) = \log_e(y).

  • Check the explicit subscript on the logarithm before exponentiating: if the base is 1010, exponentiate with base 1010 first.

Rewriting Exponential Bases

  • Any positive base bb converts to natural base ee using bx=(eln⁡(b))x=exln⁡(b)b^x = \left(e^{\ln(b)}\right)^x = e^{x\ln(b)}.

  • Substitute 10=eln⁡(10)10 = e^{\ln(10)} into 10−A10^{-A} to introduce the natural logarithm factor ln⁡(10)\ln(10) into the exponent.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from the isolated form: log⁡10(II0)=−A\log_{10}\left(\frac{I}{I_0}\right) = -A

  2. Apply base-1010 exponentiation to both sides: II0=10−A\frac{I}{I_0} = 10^{-A}

  3. Multiply both sides by I0I_0: I=I0⋅10−AI = I_0 \cdot 10^{-A}

  4. Rewrite base 1010 in terms of base ee using 10=eln⁡(10)10 = e^{\ln(10)}: I=I0⋅(eln⁡(10))−A=I0⋅e−(ln⁡(10))AI = I_0 \cdot \left(e^{\ln(10)}\right)^{-A} = I_0 \cdot e^{-(\ln(10))A}

  5. Compare with I=I0⋅e−kAI = I_0 \cdot e^{-kA} to read off kk.

Final answer

I = I_0 ⋅\cdot e^{-(\ln(10))A},  k = \ln(10)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice distinguishing when to exponentiate with base 1010 versus base ee.

  • Review the base conversion identity bu=euln⁡(b)b^u = e^{u \ln(b)} for rewriting exponential models.

Lock it in

Question

How do you rewrite the equation y=−log⁡b(x)y = -\log_b(x) in exponential form?

Answer

x=b−yx = b^{-y}. First isolate the log by multiplying by −1-1 to get log⁡b(x)=−y\log_b(x) = -y, then apply the definition of a logarithm.

Similar question

01

Solve the equation D=−log⁡10(PP0)D = -\log_{10}\left(\frac{P}{P_0}\right) for PP, and express PP in the form P=P0⋅e−cDP = P_0 \cdot e^{-c D}. Find the exact value of cc.

Answer

P=P0⋅e−(ln⁡(10))DP = P_0 \cdot e^{-(\ln(10))D}, where c=ln⁡(10)c = \ln(10). Steps: −log⁡10(P/P0)=D  ⟹  log⁡10(P/P0)=−D  ⟹  P/P0=10−D-\log_{10}(P/P_0) = D \implies \log_{10}(P/P_0) = -D \implies P/P_0 = 10^{-D}. Using 10=eln⁡(10)10 = e^{\ln(10)} gives P=P0⋅e−(ln⁡(10))DP = P_0 \cdot e^{-(\ln(10))D}.

02

In chemistry, pH=−log⁡10[HX+]\text{pH} = -\log_{10}[\ce{H+}]. Express the hydrogen ion concentration [HX+][\ce{H+}] in the form [HX+]=e−m⋅pH[\ce{H+}] = e^{-m \cdot \text{pH}}. Determine the exact constant mm.

Answer

[HX+]=e−(ln⁡(10))pH[\ce{H+}] = e^{-(\ln(10))\text{pH}}, where m=ln⁡(10)m = \ln(10). Steps: log⁡10[HX+]=−pH  ⟹  [HX+]=10−pH\log_{10}[\ce{H+}] = -\text{pH} \implies [\ce{H+}] = 10^{-\text{pH}}. Since 10=eln⁡(10)10 = e^{\ln(10)}, [HX+]=(eln⁡(10))−pH=e−(ln⁡(10))pH[\ce{H+}] = (e^{\ln(10)})^{-\text{pH}} = e^{-(\ln(10))\text{pH}}, giving m=ln⁡(10)m = \ln(10).

03

An acoustic attenuation formula is given by α=−12log⁡5(pp0)\alpha = -\frac{1}{2}\log_5\left(\frac{p}{p_0}\right). Solve for the pressure ratio pp0\frac{p}{p_0} and rewrite it as e−kαe^{-k\alpha}. Find the exact value of kk.

Answer

pp0=e−(2ln⁡(5))α\frac{p}{p_0} = e^{-(2\ln(5))\alpha}, where k=2ln⁡(5)=ln⁡(25)k = 2\ln(5) = \ln(25). Steps: Multiply by −2-2 to get log⁡5(p/p0)=−2α  ⟹  p/p0=5−2α\log_5(p/p_0) = -2\alpha \implies p/p_0 = 5^{-2\alpha}. Substitute 5=eln⁡(5)5 = e^{\ln(5)} to get p/p0=(eln⁡(5))−2α=e−(2ln⁡(5))αp/p_0 = (e^{\ln(5)})^{-2\alpha} = e^{-(2\ln(5))\alpha}.

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