Why Exponential and Logarithmic Functions Are Useful to Linearize
Exponential and logarithmic functions can change rapidly, but near a chosen input their graphs can be approximated by tangent lines. This gives a convenient way to estimate nearby values using only the function value and slope at one point.
The most useful base points are often where the function value and derivative are simple. For an exponential, \(x=0\) or \(x=1\) is often convenient. For a logarithm, \(x=1\) is especially useful because \(\ln 1=0\) and the derivative of \(\ln x\) there is \(1\).
Throughout, a linearization at \(x=a\) means the tangent-line function \(L(x)=f(a)+f'(a)(x-a)\). It estimates \(f(x)\) when \(x\) is close to \(a\); it is not generally equal to \(f(x)\) away from the base point.
Linearizing Exponential Functions
For a positive base \(b\), the derivative of \(b^x\) is \((\ln b)b^x\). Thus, at a base point \(x=a\), the function value is \(b^a\) and the tangent slope is \((\ln b)b^a\). Substituting those quantities into the definition gives the general formula.
When \(b=1\), \(b^x=1\) is constant, so its derivative and linearization slope are both zero. When \(b\ne1\), the slope depends on \(\ln b\), not just on the value \(b^a\). In particular, for the natural exponential function, \(\ln e=1\), which makes its derivative especially simple.
Worked Example: Estimate a Natural Exponential Near Zero
Use the linearization of \(f(x)=e^x\) at \(a=0\) to estimate \(e^{0.04}\).
First find the function value and slope at the base point:
Therefore the linearization is
Substitute \(x=0.04\) into the tangent-line function:
Check the tangent data: \(L(0)=1=f(0)\), and \(L'(x)=1\), so \(L'(0)=1=f'(0)\). Thus the line has the correct value and slope at the base point. This check confirms the linearization; it does not claim that \(e^{0.04}\) equals \(1.04\) exactly.
For a base other than \(e\), the logarithm factor is essential. If \(b>1\), then \(\ln b>0\), and the function increases. If \(0<b<1\), then \(\ln b<0\), and it decreases. Both cases are covered by the same formula.
Worked Example: Estimate a Power of Two Near One
Use a linearization of \(f(x)=2^x\) at \(a=1\) to estimate \(2^{1.02}\).
Evaluate the function and derivative at \(1\):
The linearization is
Now substitute \(x=1.02\). The change from the base point is \(1.02-1=0.02\), so
Using \(\ln 2\approx0.6931\), rounded to four decimal places for this calculation, gives
Check at the base point: \(L(1)=2=f(1)\). Differentiate the line to get \(L'(x)=2\ln2\), so \(L'(1)=2\ln2=f'(1)\). The estimate is therefore based on the correct tangent line.
Linearizing Logarithmic Functions
The natural logarithm is defined for positive inputs, and its derivative is \(\frac{1}{x}\). For \(f(x)=\ln x\), the function value at a positive base point \(a\) is \(\ln a\), while the slope there is \(\frac{1}{a}\). Therefore, provided \(a>0\),
This formula is useful when the target input is close to a positive number whose logarithm is known. At \(a=1\), it simplifies because \(\ln1=0\): the linearization is \(L(x)=x-1\).
Worked Example: Estimate a Natural Logarithm Near One
Use the linearization of \(f(x)=\ln x\) at \(a=1\) to estimate \(\ln(0.96)\).
At the base point,
Thus
Since \(0.96\) is positive and close to \(1\), substitute it into this line:
Check: \(L(1)=0=\ln1\), and \(L'(x)=1\), so \(L'(1)=1=(\ln x)'|_{x=1}\). The input \(0.96\) is in the domain of the logarithm, as required.
For a logarithm to a different base, use \(\log_b x=\frac{\ln x}{\ln b}\), where \(b>0\) and \(b\ne1\). Differentiating gives \((\log_b x)'=\frac{1}{x\ln b}\), so the linearization at \(a>0\) is
The restrictions matter: the input \(x\) and the base point \(a\) must be positive, and the logarithm base must be positive and different from \(1\). A base of \(1\) does not define a logarithm function.
Why the Tangent-Line Formula Works
The linearization is not an arbitrary line chosen because it looks close to the graph. Its value and slope are fixed by the function at the base point. The following derivation verifies both conditions and shows why the tangent line is the appropriate first-order model.
Set the line's value at \(a\) equal to \(f(a)\). A line through \((a,f(a))\) with slope \(m\) has equation \(L(x)=f(a)+m(x-a)\).
The derivative of this line is the constant \(L'(x)=m\). To make its slope agree with the function's slope at \(a\), choose \(m=f'(a)\). This yields \(L(x)=f(a)+f'(a)(x-a)\).
For \(f(x)=b^x\), \(f(a)=b^a\) and \(f'(a)=(\ln b)b^a\). Substitution gives \(L(x)=b^a+(\ln b)b^a(x-a)\).
For \(f(x)=\ln x\), \(f(a)=\ln a\) and \(f'(a)=1/a\), with \(a>0\). Substitution gives \(L(x)=\ln a+(x-a)/a\).
This construction also gives uniqueness: any line \(M(x)=c+m(x-a)\) that matches the function value at \(a\) must have \(c=f(a)\), and if it matches the derivative there, it must have \(m=f'(a)\). Thus it must be the linearization.
Concavity and the Direction of the Estimate
The second derivative can indicate whether a tangent-line estimate is above or below the graph. For \(b>0\),
This second derivative is positive if \(b>0\) and \(b\ne1\), and it is zero if \(b=1\). Therefore, when \(b\ne1\), \(b^x\) is strictly concave up and its tangent lines lie below the graph. When \(b=1\), the function is the constant \(1\), and its tangent line coincides with the graph. For the natural logarithm, \((\ln x)''=-1/x^2<0\) for \(x>0\), so its tangent lines lie above its graph.
These observations describe the direction of the estimate on the relevant domain, but they do not by themselves measure how large the error is. The estimate is generally most useful for inputs close to the chosen base point.
Common Mistake: Omitting the Logarithm Factor
A frequent error is to assume that the derivative of \(2^x\) is simply \(2^x\). That rule is true for \(e^x\), but not for every exponential base. For example, the incorrect calculation at \(x=0\) would use slope \(2^0=1\) and give \(2^{0.1}\approx1.1\). The correct derivative is \((\ln2)2^x\). At zero, the slope is \(\ln2\), so the correct linearization and estimate are
To check the correction, \(L(0)=1=2^0\), and \(L'(0)=\ln2=(2^x)'|_{x=0}\). The missing factor \(\ln2\) changes the tangent slope, so omitting it produces the wrong line.
Choosing and Checking a Useful Linearization
For an estimate, choose a base point near the target input where both the function value and derivative are manageable. Then calculate the change \(x-a\) carefully. Keeping that change visible helps prevent sign errors, especially when estimating logarithms at inputs below \(1\).
- For \(e^x\) near zero, use \(e^x\approx1+x\).
- For \(b^x\) near \(a\), use \(b^a+(\ln b)b^a(x-a)\).
- For \(\ln x\) near \(a>0\), use \(\ln a+(x-a)/a\).
- For \(\log_b x\), include the factor \(1/\ln b\) in the slope.
After calculating, check that the line gives the correct function value and derivative at its base point. Also verify that the logarithm's input stays positive and that any logarithm base satisfies its restrictions. These checks do not make an approximation exact; they help ensure that the correct local model was used.
Check Your Understanding
Use the formulas and checks in this tutorial to answer the following questions.
- Find the linearization of \(f(x)=e^x\) at \(a=1\), then use it to estimate \(e^{1.03}\).
- Find the linearization of \(f(x)=3^x\) at \(a=0\). Which factor in its slope would be missing if you incorrectly treated it like \(e^x\)?
- Use the linearization of \(\ln x\) at \(a=2\) to estimate \(\ln(2.1)\).
- State the domain restrictions for the linearization of \(\log_b x\) at \(a\).
- For \(b>0\), explain what happens to the concavity and tangent-line position when \(b=1\) compared with when \(b\ne1\).
