Why Choose a Base Point?
Exponential and logarithmic functions can be awkward to evaluate by hand. When the input is close to a value where the function is easy to calculate, a tangent line gives a useful estimate. The estimate is local: it is most reliable when the target input is near the chosen base point.
For exponential functions, convenient base points often make the output a familiar number. For logarithms, a base point such as \(1\) is useful because \(\ln 1=0\). A good estimate also includes a check: compare it with a numerical value when available, or use a bound on the second derivative to guarantee that the error cannot exceed a stated amount.
The Linear Estimate
Let \(f\) be differentiable at a base point \(a\). The tangent line to its graph at \(x=a\) has value \(f(a)\) there and slope \(f'(a)\). Its value at a nearby input \(x\) is the linear estimate
The base point is \(a\), and \(x-a\) is the change in input. The estimate uses the function value and slope at \(a\), not at the target \(x\). The closer \(x\) is to \(a\), the more useful the tangent line is likely to be, although closeness alone does not give a numerical error guarantee.
The formula follows directly from the definition of the derivative. For a small change \(h\), differentiability at \(a\) means that the change in function value, divided by \(h\), approaches \(f'(a)\):
This says that for small \(h\), the change \(f(a+h)-f(a)\) is approximately \(f'(a)h\). Adding \(f(a)\) to both sides of this approximate relationship gives \(f(a+h)\approx f(a)+f'(a)h\). Setting \(x=a+h\) gives the linearization formula. The approximation works because the graph is increasingly well represented by its tangent line as the input approaches \(a\).
Formulas for Exponentials and Logarithms
For \(f(x)=e^x\), the derivative is \(f'(x)=e^x\). For an exponential with base \(b>0\), \(b\ne 1\), the derivative is \(b^x\ln b\). The factor \(\ln b\) matters whenever the base is not \(e\). For the natural logarithm, \(f(x)=\ln x\), the domain is \(x>0\) and \(f'(x)=1/x\).
These are linear estimates about the chosen base point \(a\). To check an estimate with a second-derivative bound, suppose \(f\) is twice differentiable throughout the interval between \(a\) and \(x\), and \(|f''(t)|\le M\) on that interval. Then the absolute error satisfies
The bound requires control of the second derivative over the whole interval, not just at the base point. It guarantees a maximum possible error; it does not say that the actual error equals the bound.
Worked Exponential Estimates
Worked Example: Estimate \(e^{0.1}\) near zero
Use \(f(x)=e^x\) and choose \(a=0\). This is convenient because \(e^0=1\), and the derivative there is also \(1\). The target input is \(x=0.1\), so \(x-a=0.1\). The linear estimate is
Thus the estimate is \(1.1\). To check it numerically, \(e^{0.1}\approx 1.105170186\), so the absolute error is approximately
For a guaranteed bound, \(f''(t)=e^t\). On \(0\le t\le 0.1\), \(e^t\le e^{0.1}\), so take \(M=e^{0.1}\approx1.105170186\). The second-derivative bound gives
The observed error, approximately \(0.005170186\), is less than the bound, approximately \(0.005525851\). Substituting the estimate into the error calculation confirms the comparison.
Worked Example: Estimate \(2^{1.1}\) near one
Set \(f(x)=2^x\) and choose \(a=1\), where \(f(1)=2\). Its derivative is \(f'(x)=2^x\ln 2\), so \(f'(1)=2\ln 2\). At \(x=1.1\), the input change is \(0.1\). Therefore
Using \(2^{1.1}\approx2.143546925\), the absolute error is
To obtain a guarantee, \(f''(t)=2^t(\ln 2)^2\). This derivative is increasing on \([1,1.1]\), so its largest value there is \(2^{1.1}(\ln 2)^2\). Using \(2^{1.1}\approx2.143546925\) and \(\ln 2\approx0.693147181\), take \(M\approx2.143546925(0.480453014)\approx1.0299\). Hence
The numerical error, approximately \(0.004917489\), is below the guaranteed bound, approximately \(0.0051495\). The estimate also illustrates why \(2^x\) must be differentiated with the factor \(\ln 2\).
Worked Logarithm Estimates
Worked Example: Estimate \(\ln(1.04)\) near one
Choose \(f(x)=\ln x\) and \(a=1\). This point is convenient because \(\ln 1=0\), and \(f'(1)=1\). The input change is \(1.04-1=0.04\), so
A numerical value is \(\ln(1.04)\approx0.039220713\). The absolute error is therefore
Here \(f''(t)=-1/t^2\). On \([1,1.04]\), \(|f''(t)|=1/t^2\le1\), so \(M=1\) works. The error bound is
The observed error, approximately \(0.000779287\), is less than \(0.0008\). Both the direct comparison and the guaranteed bound support the estimate.
Worked Example: Estimate \(\ln(3.1)\) near three
The target \(3.1\) is close to \(a=3\), where the function value \(\ln 3\) can be used as a known reference. For \(f(x)=\ln x\), the slope at \(3\) is \(f'(3)=1/3\). The input change is \(3.1-3=0.1\), giving
Since \(\ln(3.1)\approx1.131402111\), the absolute error is
On the interval \([3,3.1]\), \(f''(t)=-1/t^2\), so \(|f''(t)|=1/t^2\le1/9\). Take \(M=1/9\). Then
The calculated error, approximately \(0.000543511\), is below the bound, approximately \(0.000555556\). This base point is useful because it keeps the change in input small while allowing the estimate to use a reference function value.
Choosing a Base Point and Checking the Result
A practical choice of base point balances two considerations: the values \(f(a)\) and \(f'(a)\) should be manageable, and the target \(x\) should be close enough to \(a\) for a local approximation to be useful. For example, \(a=0\) simplifies \(e^x\), while \(a=1\) simplifies \(\ln x\). For \(b^x\), an integer base point may make \(b^a\) easy to evaluate.
Once the estimate is calculated, check it in a disciplined order:
Identify \(f\), \(a\), and the target \(x\). Confirm that the interval between \(a\) and \(x\) stays in the function's domain.
Use \(f(a)\) and \(f'(a)\). For \(b^x\), include the factor \(\ln b\) in the derivative.
Substitute into \(L(x)=f(a)+f'(a)(x-a)\), keeping the input change \(x-a\) signed.
If a numerical function value is available, calculate the absolute difference. For a guarantee, bound \(|f''|\) across the entire interval and apply the second-derivative bound.
When comparing a numerical estimate with an error bound, keep enough digits during intermediate calculations to avoid distorting the comparison. Round only the final reported values, and state the precision used. The examples above report numerical values to nine decimal places; the error bounds are displayed to a precision that makes their comparison clear.
A Common Derivative Error
A frequent mistake is to differentiate \(b^x\) as though its derivative were just \(b^x\). That rule is correct for \(e^x\), but not for a general positive base \(b\ne1\). For instance, omitting \(\ln 2\) when estimating \(2^{1.1}\) at \(a=1\) would give the incorrect calculation
This uses the incorrect slope \(2\). The correct derivative at \(a=1\) is \(2\ln 2\), so the corrected calculation is \(2+2\ln(2)(0.1)\approx2.138629436\). Comparing with \(2^{1.1}\approx2.143546925\), the corrected absolute error is approximately \(0.004917489\). Substituting the correct slope into the tangent-line formula fixes the error.
Another important check is the domain: \(\ln x\) is defined only for \(x>0\). A logarithm estimate must use a positive base point, and the interval between the base point and target must remain positive for the second-derivative bound used here.
Why Error Checks Matter
A linear estimate is not an exact evaluation. Its value is that it replaces a difficult nearby calculation with function and derivative values at a convenient point. An error check tells you how much trust to place in the result. A numerical comparison reveals the actual discrepancy when a reference value is available; a second-derivative bound guarantees an upper limit under its stated conditions.
These methods are useful in applied calculations whenever an exponential or logarithmic quantity changes by a small amount. The key habits are to choose a suitable base point, use the derivative that matches the function, track the input change carefully, and distinguish an observed error from a guaranteed bound.
Check Your Understanding
For each question, identify the base point and explain how you would check the estimate.
- Write the linearization of \(e^x\) at \(a=0\), then use it to estimate \(e^{0.06}\).
- What factor must appear in the derivative of \(5^x\), and why is it absent when differentiating \(e^x\)?
- Use the linearization of \(\ln x\) at \(a=1\) to estimate \(\ln(0.97)\).
- What information about \(f''\) is required to apply the second-derivative error bound between \(a\) and \(x\)?
- Explain the difference between a numerical absolute error and a guaranteed error bound.
