An exponential has its variable in the exponent, and its difference quotient factors exactly: bᵗ times a quotient that does not contain t. That second factor is a column of rounded numbers, and the lesson states, without proving it, that the column settles on a constant. The rate of bᵗ is that constant times bᵗ, and e is the base whose constant is 1.
Core ideas
Three ideas: the exact factoring, what the column of the remaining factor claims, and why this is not the power rule.
The law of exponents from Algebra's Exponential and Logarithmic Functions says b^(a + h) = b^a·b^h. So b^(a + h) − b^a = b^a·(b^h − 1), and dividing by h gives b^a·(b^h − 1)/h. The second factor contains the step h and the base b, and nothing about the place a.
At a = 0, b^a = 1, so the quotient there is the second factor alone. That one column, for halving h, is the same at every place after multiplying by b^a. The lesson states that the column settles on one constant. A table of rounded rows illustrates this and does not prove it.
The power rule differentiates tⁿ, with the variable in the base and the exponent fixed. In 2ᵗ it is the other way round, and the rule does not apply. The rate of 2ᵗ is proportional to 2ᵗ itself, with a constant of proportion, and e is the base for which that constant is 1.
The material
The factoring, the column for the base 2, what is exact and what is rounded, the other bases, and the failure of the power rule.
Apart from 1/t, everything before this lesson was a polynomial, whose quotient is a polynomial in the step and comes out exact. An exponential is not a polynomial, and for most bases and steps its quotient is no longer a fraction, so for the first time in the course some figures are rounded.
That is exact algebra and nothing is dropped. Every quotient of bᵗ at every place is the base raised to the place, times the quotient at the place 0. Everything depends on the second factor, so look at it for b = 2.
At h = 1 it is (2 − 1)/1 = 1, exactly, because the base is a whole number and the step is 1. At h = 1/2 it is (√2 − 1)/(1/2) = 2(√2 − 1), which has no finite decimal, and the lab prints ≈ 0.828427. From the second row on, every figure in the table is rounded to six significant figures and carries the sign ≈.
For each base b > 0 the quotients (b^h − 1)/h for halving h head for a number written ln b. Then the rate of bᵗ at every t is ln b times bᵗ. The base e ≈ 2.71828 is the one with ln e = 1, so (eᵗ)′ = eᵗ.
This is stated and not proved. The factoring above is proved; that the rounded column settles on one number is what the table suggests and the lab shows for each base, and no table of seven rows establishes it for every step. For b = 2 the column heads for ln 2 ≈ 0.693147, the way the quotients of t² headed for 2.
For b = 3 the first quotient is (3 − 1)/1 = 2, exact, and the column heads for ln 3 ≈ 1.09861. For b = e, entered in the lab as e, every quotient is rounded, since e itself is not a fraction: the first is e − 1 ≈ 1.71828, and the column heads for 1. That is the definition of the number: of all the bases, e is the one for which the exponential's rate is the exponential itself.
At a = 3, with b = 2, the quotient at h = 1 is (16 − 8)/1 = 8, exactly 2³ times the first entry of the column above. Every later row is 8 times the matching row of that column, so the rate of 2ᵗ at 3 is 8·ln 2 ≈ 5.54518, rounded. The lab's ratio tile divides the last quotient by 2³ and gets back the column at 0.
Applying the power rule to 2ᵗ would give t·2^(t − 1). At t = 0 that is 0, but the quotients of 2ᵗ at 0 head for ≈ 0.693147. The power rule belongs to a variable base and a fixed exponent. When the variable is in the exponent, the rate is proportional to the function.
A quantity whose rate is a constant multiple of itself is exactly what the equation y′ = k·y says, and this lesson has shown that exponentials have the property. That equation opens the second half of this Subject, in Separable Equations, Growth and Decay.
Interactive
Rounded quotients, labelled, and what they approach
Method
Five moves. The first is exact algebra and the rest read a rounded table, so keep the two apart.
Write b^(a + h) as b^a·b^h, take out b^a, and the quotient is b^a·(b^h − 1)/h. This step is exact.
At a = 0 the factor b^a is 1, so the quotient is (b^h − 1)/h. Halve the step row by row. For the bases 2 and 3 only the first entry is a fraction, and the lab prints every later one rounded.
Every figure with ≈ is a rounded decimal of a number with no finite decimal. Never quote one as exact, and never read the last row as the constant itself.
The quotient at a is b^a times the column at 0, so the constant the column heads for is the same at every place.
For b = 2 it is ln 2 ≈ 0.693147, for b = 3 it is ln 3 ≈ 1.09861, and for b = e it is exactly 1.
The first row is exact. Every row after it is rounded to six significant figures, and the lab says so beside it.
The factoring is exact, so the second block is the first multiplied by 8 row for row. The constant is claimed, and the lab shows the column sinking toward it.
For a rehearsal, take b = 3 at a = 0: the first entry is 2, and the column should head for ≈ 1.09861, which is ln 3.
You should be able to write the quotient of bᵗ as b^a·(b^h − 1)/h, say which entries of the column are exact and which are rounded, name the constant for a given base, and say why the power rule does not apply.
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