No number shows up more often in the rest of this course than e \approx 2.71828\ldots. No pair of functions undoes each other more often than e^x and \ln x. The reason e earns that spotlight is itself a calculus fact: e^x is the one exponential function whose steepness at every point equals its own height, with no extra constant multiplied in. But the algebra of exponents and logs underneath that fact needs to already be automatic.
An exponential function is compounding. Whatever the output is right now, the next step multiplies it by the same fixed factor, over and over. Logarithms ask the un-doing question: this quantity compounded by a fixed factor and ended up here, so how many compounding steps did that take? Growth happening every single instant, not once a year, is what e measures: the compounding rate where that process works out to one clean, describable constant. It's the natural ceiling on pure compounding. That sense gets made completely precise once the tools for measuring instantaneous change exist.
An exponential function b^x (base b>0, b\neq 1) multiplies by the fixed factor b once for every unit increase in x. The logarithm \log_b x is b's inverse: the exponent you'd have to raise b to in order to get x. So \log_b x = y means exactly b^y = x. Every log property is just that inverse relationship translated into exponent language. For instance, \log_b(mn)=\log_b m+\log_b n mirrors b^{y_1}b^{y_2}=b^{y_1+y_2}, since multiplying outputs corresponds to adding exponents. The constant e \approx 2.71828\ldots is a specific, irrational base singled out for one reason: e^x is the one exponential function whose steepness at every point equals its own height, with no extra constant multiplied in. That fact is proved once calculus's rate-of-change tools exist, not here, but it's why e earns a permanent name rather than staying "just some number." \ln x is just \log_e x, the natural logarithm, base e. It is by far the most common base once calculus starts: \frac{d}{dx}\ln x = \frac{1}{x}, and no other base gives that clean a result.
Formula
Exponential Function
also known as: the general exponential model
Recalled from College Algebra · Exponential functions — first taught there.
An exponential function with base b has the form f(x) = a \cdot b^x, \qquad b > 0,\; b \neq 1,\; a \neq 0, where a = f(0) is the initial value. It models growth when b > 1 and decay when 0 < b < 1.
Formula
Definition of a Logarithm
also known as: log-exponential equivalence
Recalled from College Algebra · Logarithms — first taught there.
For b > 0, b \neq 1, and x > 0, \log_b x = y \quad \Longleftrightarrow \quad b^y = x. That is, \log_b x is the exponent to which b must be raised to produce x.
For b > 0, b \neq 1, positive reals M, N, and any real p:
- Product rule: \log_b(MN) = \log_b M + \log_b N
- Quotient rule: \log_b\frac{M}{N} = \log_b M - \log_b N
- Power rule: \log_b(M^p) = p \log_b M
Growth vs. decay
Slide k through 0 — positive means growth, negative means decay, and k=0 is a flat line.
Loading interactive…
Where is ln(x) - 1 positive?
Click each interval — \ln x - 1 crosses zero at x = e \approx 2.71828, and stays positive only to the right of it.
Loading interactive…
Worked example
Expanding a log expression
-
1
Expand \ln\!\left(\dfrac{x^3\sqrt{x+1}}{5}\right) as far as possible.
-
2
Rewrite the square root as a 1/2 power first, so every piece is a clean product/quotient/power: \ln\!\left(\dfrac{x^3(x+1)^{1/2}}{5}\right).
-
3log of a quotient splits into a difference of logs
Apply the quotient property of logs: \ln\!\big(x^3(x+1)^{1/2}\big) - \ln 5.
log of a quotient splits into a difference of logs
-
4
Apply the product property of logs on the remaining numerator: \ln(x^3) + \ln\!\big((x+1)^{1/2}\big) - \ln 5.
-
5
Apply the power property of logs to each term: 3\ln x + \tfrac{1}{2}\ln(x+1) - \ln 5.
Loading interactive…
Where is ln(x-2) even defined?
Drag the probe — the log only exists to the right of x = 2, exactly where the argument x-2 is positive.
Loading interactive…
Where this goes next
How fast e^x changes, two chapters ahead
Every other exponential's rate-of-change formula (\frac{d}{dx}b^x = b^x \ln b) carries a leftover \ln b factor. When b = e, \ln e = 1 makes that factor disappear entirely. That's the reason e gets singled out: it's the one base whose rate of change doesn't pick up any extra baggage.
Check your understanding (counts toward your progress)
For P(t) = 500e^{0.04t}, find P(0).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
P(0)=500e^{0.04(0)}=500e^0=500(1)=500.
Correct answer
Also accepted:
Why
Expanding \log_b\left(\dfrac{x^{5}y^{5}}{z^{8}}\right) as a sum/difference of logs, what is the coefficient of \log_b x?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\log_b\left(\dfrac{x^{5}y^{5}}{z^{8}}\right)=5\log_b x+5\log_b y-8\log_b z, so the coefficient of \log_b x is 5.
Correct answer
Also accepted:
Why
Each factor's exponent becomes that term's coefficient; a denominator factor's term is subtracted.
Practice: Exponentials and logs, working both directions (3 problems · doesn’t affect your progress)
Which of these correctly expands \log_b(x^2y^3)?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Adding the exponents (2+3=5) and dropping them onto a combined xy loses the separate powers -- the power rule has to be applied to each factor before combining.
This turns the exponents into coefficients multiplying x and y themselves, which isn't what the power rule does -- the exponents belong on the logs, not inside them.
Multiplying the two exponents (2 times 3 = 6) has no basis in any log property -- the product and power rules only ever add exponents, never multiply them.
Why
The product rule splits \log_b(x^2y^3) into \log_b(x^2)+\log_b(y^3), and the power rule moves each exponent out front, giving 2\log_b x+3\log_b y.
Apply the product rule first (splitting x^2 y^3 into two logs), then the power rule on each piece separately.
2\log_b x + 3\log_b y
Evaluate e^{\ln 7}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
e^x and \ln x are inverse functions, so e^{\ln 7} undoes back to exactly 7.
Correct answer
Also accepted:
Why
e^x and ln(x) undo each other completely -- e raised to ln(anything) just returns that anything.
Without computing an exact decimal, is \log_{10}(950) closer to 2 or to 3?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
950 is only 50 away from 1000 but 850 away from 100 -- it sits far closer to 10^3.
Halfway between 2 and 3 on this logarithmic scale would be near \log_{10}(316), not 950 -- 950 is much higher than that.
Bracketing the number between two nearby powers of 10 gives a solid estimate without any computation at all.
Why
10^2=100 and 10^3=1000; 950 is only 50 away from 1000 but 850 away from 100, so \log_{10}(950) is much closer to 3.
Compare how far 950 sits from 100 (=10^2) versus from 1000 (=10^3).
Closer to 3, since 950 is much nearer to 10^3=1000 than to 10^2=100
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Evaluate \log_2(32) + \log_2\left(\frac{1}{8}\right).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\log_2 32 = 5 (since 2^5=32) and \log_2(1/8) = -3 (since 2^{-3}=1/8). Adding: 5 + (-3) = 2.
Correct answer
Also accepted:
Why
Evaluate each logarithm separately first — \log_2 32 and \log_2(1/8).
Then add the two values together.
From College Algebra
Evaluate \log_5 \frac{1}{125}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
'5 to what power gives \frac{1}{125}?' Since 5^3=125, we need 5^{-3}=\frac{1}{125}, so the answer is -3.
Correct answer
Also accepted:
Why
Rewrite \frac{1}{125} as a power of 5 with a negative exponent.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Evaluate \log_2(4^3 \cdot 2) using log properties.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Product rule then power rule: \log_2(4^3\cdot 2) = \log_2(4^3) + \log_2 2 = 3\log_2 4 + 1 = 3(2) + 1 = 7.
Correct answer
Also accepted:
Why
Split the argument with the product rule into \log_2(4^3) + \log_2 2, then apply the power rule to the first term.