Introduction
Exponential and logarithmic functions appear in models of growth, decay, and change of scale. Their values can be difficult to calculate exactly, but near a suitable point they can often be replaced by a simple line.
A linearization uses the function’s value and slope at one point to form a nearby estimate. For exponential and logarithmic functions, the point \(0\) is especially useful: \(e^0=1\), and \(\ln(1+0)=0\). These facts make the resulting formulas particularly simple.
The Linearization Formula
If \(f\) is differentiable at \(a\), its linearization at \(a\) is the line with the same value and slope as \(f\) at that point. Its formula is
The input \(a\) is called the base point. The model agrees with the function at that point: \(L(a)=f(a)\). Its slope is \(f'(a)\), so it also matches the function’s instantaneous rate of change there. When \(x\) is close to \(a\), \(L(x)\) is used as an estimate for \(f(x)\).
Why Matching the Value and Slope Gives This Formula
A line has the form \(L(x)=mx+b\), where \(m\) is its slope. To make it match \(f\) at \(a\), require \(L(a)=f(a)\). To make its slope match the function’s slope there, require \(m=f'(a)\). Substituting \(x=a\) into the line gives
With \(m=f'(a)\), this determines the line completely:
The formula is not an arbitrary convenient expression: it is the unique line that matches both the function’s value and its derivative at the base point. The derivative describes the local slope, which is why this line is useful close to \(a\). The closer the input is to the base point, the more appropriate it is to use the function’s behavior there as a model.
Exponential Linearizations
The derivative of \(e^x\) is \(e^x\). At \(a=0\), its value and derivative are both \(1\). Therefore,
Thus, near zero, \(e^x\) is estimated by \(1+x\). More generally, for a positive base \(c\), the derivative of \(c^x\) is \(c^x\ln c\). At zero, \(c^0=1\), so the linearization is
The condition \(c>0\) is required for the real-valued exponential function \(c^x\) and its derivative formula. When \(c=e\), \(\ln c=1\), giving the simpler model \(1+x\).
Worked Example: Estimate \(e^{0.08}\)
Use the linearization of \(e^x\) at zero. Here \(f(x)=e^x\), \(f(0)=1\), and \(f'(0)=e^0=1\). Hence
Substitute \(x=0.08\):
The model’s arithmetic checks directly: \(L(0.08)=1.08\), and at the base point \(L(0)=1=e^0\). The actual function value is \(e^{0.08}\approx1.083287\), so the linear estimate is \(1.08\), or \(1.0800\) to four decimal places. The actual value rounds to \(1.0833\) to four decimal places.
Worked Example: Estimate \(2^{0.05}\)
Set \(f(x)=2^x\). Its value at zero is \(f(0)=1\), and its derivative is \(f'(x)=2^x\ln 2\). Therefore \(f'(0)=\ln 2\), and the linearization at zero is
Using \(\ln 2\approx0.6931471806\), evaluate the model at \(x=0.05\):
Thus the linear estimate is \(1.0346573590\), or \(1.0347\) rounded to four decimal places. The actual value is \(2^{0.05}=e^{0.05\ln 2}\approx1.0352649238\), which rounds to \(1.0353\) to four decimal places. These are different quantities: \(1.0347\) is the rounded linear estimate, not the actual value rounded to four decimal places.
As a check on the model, \(L(0)=1\), matching \(2^0=1\), and its slope is \(\ln 2\), matching \(f'(0)\). The substitution into the formula also gives the displayed estimate.
Logarithmic Linearizations
The natural logarithm \(\ln x\) is defined for \(x>0\), and its derivative is \(1/x\). At the convenient base point \(a=1\), the function value is \(\ln 1=0\) and the slope is \(1/1=1\). Its linearization is consequently
This model estimates \(\ln x\) when \(x\) is close to \(1\). Writing an input near \(1\) as \(1+h\) gives the equivalent estimate \(\ln(1+h)\approx h\), provided \(1+h>0\).
Alternatively, define \(g(x)=\ln(1+x)\). Its real-valued domain is \(x>-1\). The derivative is \(g'(x)=1/(1+x)\), so \(g(0)=0\) and \(g'(0)=1\). Its linearization at zero is
Therefore, for \(x\) near zero with \(x>-1\), use \(\ln(1+x)\approx x\). The domain condition matters: if \(x\leq -1\), then \(1+x\leq0\), and the real logarithm in this expression is undefined.
Worked Example: Estimate \(\ln(1.06)\)
Write \(1.06=1+0.06\), so the input to \(g(x)=\ln(1+x)\) is \(x=0.06\). The base point is zero, where \(g(0)=0\) and \(g'(0)=1\). Thus \(L(x)=x\), and
The input satisfies the domain requirement \(0.06>-1\). The model substitution checks as \(L(0.06)=0.06\), and at its base point \(L(0)=0=\ln(1)\). The actual value is \(\ln(1.06)\approx0.0582689081\), so the estimate is \(0.0600\) to four decimal places, while the actual value is \(0.0583\) to four decimal places.
Worked Example: Estimate \(\ln(0.97)\)
The input \(0.97\) is close to \(1\), so use the linearization of \(f(x)=\ln x\) at \(a=1\). We have \(f(1)=0\) and \(f'(1)=1\), which gives
Substitute \(x=0.97\):
This is within the domain \(x>0\) of the logarithm. The model gives \(L(1)=0=\ln 1\), and \(L(0.97)=-0.03\) checks the arithmetic. The actual value is \(\ln(0.97)\approx-0.0304592075\), which rounds to \(-0.0305\) to four decimal places; the linear estimate \(-0.0300\) is not the same rounded value.
Choosing and Using a Base Point
The base point should make the function value and derivative easy to evaluate, while being close to the input of interest. For these functions, zero is natural for \(e^x\), \(c^x\), and \(\ln(1+x)\); one is natural for \(\ln x\). The equivalent logarithm models \(x\) near zero and \(x-1\) near one reflect the same tangent line written with different input notation.
Prefer a point where the function and derivative are straightforward to calculate.
Evaluate \(f(a)\) and \(f'(a)\), checking that the input lies in the function’s domain.
Use \(L(x)=f(a)+f'(a)(x-a)\), then substitute the desired input into \(L\).
Report it as a linear estimate, not as an exact function value.
A Common Mistake: Confusing the Model with the Function
A linearization gives an estimate; it does not redefine the function or guarantee that the estimate equals the actual value. For example, from the linearization \(L(x)=1+x\) for \(e^x\), a mistaken conclusion would be
The equality is wrong. The correct statement is
The same care is needed when rounding. Round the linear estimate as an estimate, and do not present it as the rounded actual value. Also check that the input is near the base point and in the function’s domain. In particular, \(\ln(1+x)\) requires \(x>-1\), while an estimate based at zero is intended for inputs near zero, not arbitrary inputs in the domain.
Summary
Exponential and logarithmic linearizations follow from the same tangent-line formula. At zero, \(e^x\) is modeled by \(1+x\), and \(c^x\) for \(c>0\) is modeled by \(1+(\ln c)x\). Near zero, \(\ln(1+x)\) is modeled by \(x\); near one, \(\ln x\) is modeled by \(x-1\). Each model matches the function’s value and slope at its base point, and is used as an approximation for nearby inputs.
Check Your Understanding
For each question, identify the base point and use the appropriate linearization. Keep the estimate distinct from the actual function value.
- Find the linearization of \(e^x\) at \(x=0\), and use it to estimate \(e^{-0.04}\).
- Find the linearization of \(3^x\) at \(x=0\). What is its slope?
- Use a logarithmic linearization to estimate \(\ln(1.03)\). State the domain condition for this model.
- Use the linearization of \(\ln x\) at \(x=1\) to estimate \(\ln(1.02)\).
- Explain why an estimate produced by a linearization should not be written as an exact equality with the function value.
