Comparing Exponential Growth Rates

Comparing Exponential Functions

An exponential function can grow quickly, but the phrase “grows faster” needs a precise meaning. A useful way to compare exponential functions is to examine how much each value is multiplied by when the input increases by one. That multiplier is the function’s successive-value ratio.

Comparisons also depend on the input. Two exponential functions may agree at one input and differ at others. A graph intersection records an input where their values are equal; it does not, by itself, say which function grows faster overall.

The Successive-Value Ratio

To see why the ratio is constant, use the exponent law \(a^{x+1}=a^x a\). Since \(C a^x\) is positive, division is valid:

Thus, each increase of one in the input multiplies the output by the same factor. If \(a=2\), each step doubles the value. If \(a=\frac{1}{2}\), each step halves it. The ratio describes a repeated multiplicative change, not the size of the function at a particular input.

For example, the function \(100(1.1)^x\) has a larger value at \(x=0\) than \(2^x\), but its successive-value ratio is \(1.1\), while the ratio for \(2^x\) is \(2\). The first function starts higher; the second multiplies by more at each step.

Comparing Different Bases

For the unscaled functions \(f(x)=a^x\) and \(g(x)=b^x\), their comparison can be made through their quotient:

The quotient shows whether \(f(x)\) is larger or smaller than \(g(x)\). If it is greater than \(1\), then \(f(x)>g(x)\); if it is less than \(1\), then \(f(x)<g(x)\). The behaviour depends on both the bases and the sign of \(x\).

To justify the comparison for positive inputs, suppose first that \(a>b\). Then \(a/b>1\), so \((a/b)^x>1\) when \(x>0\). Therefore \(a^x>b^x\). For negative inputs, write \(x=-u\) with \(u>0\). Then

because \((a/b)^u>1\). Hence \(a^x<b^x\) for negative \(x\). Finally, at \(x=0\), both values are \(1\). Since the quotient differs from \(1\) for every nonzero \(x\), the graphs have no other intersection. If instead \(a<b\), the same argument with the roles reversed gives the stated comparisons.

This result includes bases between \(0\) and \(1\). For instance, when \(0<a<b<1\), the larger base \(b\) gives the larger value for positive inputs, while the smaller base \(a\) gives the larger value for negative inputs. These functions are decay functions as the input increases, but they can still be compared by their ratios.

Finding an Exact Intersection

When the functions have different starting values or shifted inputs, their graphs may intersect somewhere other than \(x=0\). An exact intersection can sometimes be found by rewriting both sides with a common base. The key fact is that, for a base greater than \(1\), equal powers have equal exponents: the function \(2^x\), for example, is strictly increasing.

Starting Values Can Change Which Function Is Larger

A positive coefficient changes the starting value without changing the successive-value ratio. For \(F(x)=C a^x\), the value at \(x=0\) is \(C\), since \(a^0=1\), while the successive-value ratio remains \(a\). Thus a function that starts higher can have a smaller ratio and eventually be overtaken by another function.

For two models \(F(x)=C a^x\) and \(G(x)=D b^x\), with positive \(C,D,a,b\), their quotient is

Comparing this quotient with \(1\) tells which model has the larger value at a given input. When the bases differ, the quotient changes strictly in one direction as \(x\) increases. Consequently, the graphs can intersect at most once. A particular intersection may or may not be easy to find exactly; the following example has an exact solution using exponent laws.

A Common Mistake: Confusing Value with Growth Rate

A common error is to compare only the values at one input and treat that comparison as a conclusion about the rates. For example, at \(n=0\), \(2^n=1\) and \(3^n=1\). The incorrect conclusion would be: “The functions have the same value here, so they grow at the same rate.” That inference is false.

The correction is to calculate successive ratios. For \(2^n\), the ratio from one integer input to the next is \(2\); for \(3^n\), it is \(3\). The functions agree at \(n=0\), but their multipliers differ, so their values differ for positive integer inputs. A single value tells where two functions meet, not how their values change from step to step.

It is also a mistake to assume that the function with the larger initial value must remain larger. The models in the previous example show why this fails: \(B(0)=12\) is greater than \(A(0)=3\), but \(A\) has the larger successive ratio and overtakes \(B\) after their intersection.

Using the Comparison

Successive ratios give a compact way to describe multiplicative change, while intersections show where two models have equal values. These ideas are useful when comparing quantities that change by repeated multiplication, such as populations, balances, or measurements recorded at regular time intervals. To make a fair comparison, keep the input step consistent: a ratio over a one-unit increase should not be compared directly with a ratio over a different length of time.

For a reliable comparison, identify both the bases and any coefficients or shifts. The base determines the successive-value ratio; coefficients affect the starting value, and shifts affect where particular values occur. When seeking an intersection, solve the equality and substitute the proposed input into both functions. When the exact intersection is not apparent from exponent laws, do not guess from a few sample values: those values can locate a possible crossing interval, but do not establish the exact intersection.