The exponential function and the number e
For a>0 and a≠1, y=a^x stays positive and multiplies by a for each unit increase in x. Its horizontal asymptote is y=0. For y=e^x, the tangent gradient equals the function value everywhere.
What a learner can do afterwards
- Describe the shape, allowed inputs, positive outputs and asymptotes of y=2^x and y=2^(-x); calculate exponential values and identify fixed multipliers
- Say what makes e different from any other base
- Use a model of the form A e^(kt) to describe growth and decay
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Compound change repeats a multiplier. For y=a^x, use a>0 and a≠1. Every real input x is allowed, and every output is positive. At x=0, y=1. Increasing x by 1 multiplies y by a. Divide the next output by the previous output to find this multiplier from values one unit apart. A base above 1 gives growth; a base between 0 and 1 gives decay.
For y=2^x, the curve rises to the right and approaches zero to the left. For y=2^(-x), it falls to the right and approaches zero to the right. Replacing x by -x reflects in the y-axis: (x,y) becomes (-x,y). Negative powers give reciprocals: 2^(-2)=1/2^2=1/4. For every allowed base a>0, a≠1, the curve y=a^x stays above its horizontal asymptote y=0 and never reaches it.
The tangent gradient is the curve's rate of change at one point. The derivative gives these gradients. For y=e^x, the gradient equals e^x everywhere. At x=0, both the value and gradient are 1. Here e≈2.718, so bases such as 3 are larger. An interval increase is the end output minus the start output; it need not equal the starting tangent gradient.
For a positive amount, use N=A e^(kt), with A>0 and elapsed time t≥0. The exponent kt means k times t. At t=0, e^0=1, so N=A. Each time unit multiplies N by e^k. Positive k gives growth; negative k gives decay toward zero; k=0 leaves N unchanged. The rate of change is k times the current amount. This is a fixed proportional rate, unlike adding or removing a fixed amount each time unit.
Evaluate M=18e^(0.1t) at t=10. First calculate the exponent: 0.1×10=1. Then M=18e^1=18e. Using e≈2.718 gives M≈48.924. Calculate the power before multiplying by 18.
For a>0 and a≠1, y=a^x stays positive. For y=e^x, the tangent gradient equals the function value. In A e^(kt), A>0 sets the start and k sets growth, decay or no change.
Lesson and practice text written for this dataset. Background sources studied: OpenStax Algebra and Trigonometry, sections 6.1 and 6.2, and Calculus Volume 1, section 6.8.
Readings and practice questions written for this dataset, with factual background studied in OpenStax. The video references identify Khan Academy materials; the videos remain their creators' work.
More to read
Checking the multiplier
For a rule known to be exponential, divide outputs at consecutive unit inputs to find its multiplier. Suppose the outputs at x=0, 1 and 2 are 4, 20 and 100. Both ratios, 20/4 and 100/20, equal 5. The rule is y=4×5^x.
The initial output and multiplier do different jobs. The initial output is the value at input zero. The multiplier tells how the output changes when the input increases by one. In y=a^x, the initial output is 1 and the multiplier is a.
The base a must be positive and different from 1 for the exponential family studied here. Inputs can be any real numbers. For example, a negative input produces a reciprocal power, not a negative output.
Matching a few observations does not prove that an unknown process will remain exponential. A model assumes that its multiplier continues over the interval where it is used.
Following an exponential curve
Read the direction of change before sketching. On y=2^x, moving one unit right doubles the output. Moving one unit left halves it. Far to the left, these positive outputs become close to zero. The horizontal asymptote is y=0.
For y=2^(-x), the direction reverses. Increasing x makes -x smaller, so the outputs decrease. This curve approaches zero on the right while remaining above the axis. Exponential decay here means that a positive amount becomes smaller; it does not mean that the outputs become negative.
The two curves are reflections in the y-axis. Reflection changes a point's horizontal coordinate from x to -x while keeping its vertical coordinate unchanged. Both curves pass through (0,1), which remains fixed under this reflection.
A finite table can help check a sketch's direction. It cannot show an entire infinite curve or prove that the curve never reaches its asymptote.
Interpreting A and k
In N=A e^(kt), A is positive and t measures elapsed time from zero. The product kt is the whole exponent. At the start, kt=0 and e^0=1, so the amount is A.
The parameter k gives the rate of change relative to the current amount. Its numerical value depends on the time unit. With a fixed time unit, positive k gives growth and negative k gives decay. Setting k=0 leaves the amount at A for every time.
Over one time unit, the amount is multiplied by e^k. Thus k is not itself the ordinary percentage change over that unit.
For a calculation, work from the exponent outward. If Q=12e^(0.5t) and t=2, the exponent is 1 and Q=12e. Evaluate the power before multiplying by 12.
These models assume a fixed proportional rate. Real conditions may change that rate. A population model can also produce fractional outputs even though an actual population contains a whole number of individuals.
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