Exponential and Logarithmic Derivative Comparison Problems

Comparing Rates at the Same Input

Exponential and logarithmic functions are closely connected, but their derivatives behave differently as the input changes. A useful comparison asks what happens at the same input: are both functions increasing or decreasing, which derivative has greater magnitude, and how quickly is each function changing relative to its own value?

To make the comparison precise, we will use a common base \(a\). The exponential function \(a^x\) is defined for every real \(x\) when \(a>0\). The logarithm \(\log_a x\) is defined only for \(x>0\), and requires \(a\ne 1\). Thus, a same-input comparison between these functions is restricted to \(x>0\).

Derivative Formulas and Sign Conditions

For \(a>0\), the derivative of \(a^x\) is \(\ln(a)a^x\). For \(a>0\), \(a\ne1\), and \(x>0\), the derivative of \(\log_a x\) is \(1/(x\ln(a))\). The natural logarithm \(\ln(a)\) is positive when \(a>1\) and negative when \(0<a<1\).

Because \(a^x>0\) and \(x>0\), the signs of these derivatives are determined by \(\ln(a)\). For \(a>1\), both derivatives are positive, so both functions are increasing. For \(0<a<1\), both derivatives are negative, so both are decreasing. The case \(a=1\) is excluded for logarithms and would give a constant exponential function rather than a meaningful logarithmic comparison.

Having derivatives with the same sign does not mean the functions change at the same rate. To compare the sizes of their rates at one input, compare the absolute values of their derivatives. At the same \(x>0\), their ratio is

This ratio is positive for every permitted \(a\) and \(x\). If it exceeds \(1\), the logarithmic function has the larger derivative magnitude at that input. If it is less than \(1\), the exponential function has the larger derivative magnitude. The formula provides a comparison, not a claim that one function always changes faster: the result depends on both the base and the input.

Why the Logarithmic Derivative Has This Form

The logarithm and exponential with the same base undo one another: \(a^{\log_a x}=x\) for \(x>0\). Differentiating this identity explains the logarithmic derivative formula. Set \(u=\log_a x\), so \(a^u=x\). By the chain rule, differentiating both sides with respect to \(x\) gives \(\ln(a)a^u u'=1\). Since \(a^u=x\), solving for \(u'\) gives the formula below.

This calculation requires \(x>0\), \(a>0\), and \(a\ne1\): these conditions ensure that \(\log_a x\) is defined and that \(\ln(a)\ne0\), so division by \(\ln(a)\) is valid.

Relative Rates at a Matched Input

For the exponential function \(f(x)=a^x\), its relative rate is constant:

For the logarithmic function \(g(x)=\log_a x\), the relative rate is defined only when \(g(x)\ne0\). Since \(\log_a x=0\) exactly when \(x=1\), it is undefined there. For \(x>0\) with \(x\ne1\), use \(g(x)=\ln(x)/\ln(a)\) to obtain

Thus the exponential relative rate depends only on the base, whereas the logarithmic relative rate depends on the input and not on the base. The logarithmic relative rate is negative for \(0<x<1\) and positive for \(x>1\). This sign concerns change relative to the function’s value: a negative relative rate can occur even when the logarithm itself is increasing, because \(\log_a x\) is negative on \(0<x<1\) when \(a>1\).

Worked Comparisons

Same Input Is Not the Same as Inverse-Related Inputs

Although \(a^x\) and \(\log_a x\) are inverse functions, comparing them at the same input is different from comparing their derivatives at inverse-related inputs. If the exponential input is \(t\), its output is \(a^t\). The corresponding logarithmic input is therefore \(a^t\), not \(t\). At those paired inputs, the derivatives are reciprocals:

Their product is \(1\). This does not say that the derivatives at a shared input \(x\) are reciprocals. For a same-input comparison, evaluate both derivative formulas at that same positive \(x\), as in the worked examples.

A Common Mistake

A frequent error is to place \(\ln(a)\) in the numerator when differentiating a logarithm. For example, the incorrect calculation \((\log_2 x)'=\ln(2)/x\) would give \(\ln(2)/2\) at \(x=2\). The correct derivative is \(1/(x\ln(2))\), giving \(1/(2\ln(2))\) there. These results differ.

The identity \(2^{\log_2 x}=x\) checks the correction. Differentiating both sides gives \(\ln(2)2^{\log_2 x}(\log_2 x)'=1\). Since \(2^{\log_2 x}=x\), this becomes \(\ln(2)x(\log_2 x)'=1\), so \((\log_2 x)'=1/(x\ln(2))\). Keep the two formulas distinct: the exponential derivative multiplies by \(\ln(a)\), while the logarithmic derivative divides by \(\ln(a)\).