Rewrite in Terms of Base e: The Natural Base, Continuous Compounding, and Change of Base

Start with 1 dollar at 100% annual interest - the setting where mathematicians first needed to rewrite in terms of base e.

Compounded once a year, it ends at $(1+1)^1 = 2.00$. Twice a year: $(1+\tfrac{1}{2})^2 = 2.25$. Four times: $(1+\tfrac{1}{4})^4 = 2.4414$. Monthly: $(1+\tfrac{1}{12})^{12} = 2.6130$. Daily: $(1+\tfrac{1}{365})^{365} = 2.7146$.

Gains shrink at every step, and no frequency breaks past 2.71828: as $n$ grows, $(1+\tfrac{1}{n})^n \to e$, and $2 < e < 3$ - the ceiling every rewrite in terms of base e leans on.

Hall and Knight's 1887 algebra says it vividly: with interest "convertible into principal every moment," the amount becomes $Pe^{rn}$. That ceiling is $e \approx 2.71828$, waiting at the end of every rewrite in terms of base e.

The chart below shows the climb - the first rewrite in terms of base e picture to keep in mind.

The number $e \approx 2.71828$ is just a number - irrational, squeezed between 2 and 3.

A logarithm with base $e$ is a natural logarithm, written $\ln x$ - the ask half of a rewrite in terms of base e. It asks: $e$ to what power gives $x$?

Two anchors come free. Since $e^0 = 1$, we get $\ln 1 = 0$. Since $e^1 = e$, we get $\ln e = 1$.

Better, $e^x$ and $\ln x$ undo each other: $\ln(e^x) = x$, and $e^{\ln x} = x$ for $x > 0$. That inverse pair is the machinery behind every rewrite in terms of base e, which leans on both directions. On the curve $y = \ln x$ below, every rewrite in terms of base e finds its anchor: height exactly 1 at $x = e$.

Take one concrete fact: $3^2 = 9$. Now rewrite in terms of base e: $e^{2\ln 3} = 9$.

Why does that work? $\ln 3 \approx 1.0986$, so $2\ln 3 \approx 2.1972$, and $e^{2.1972} \approx 9$. Same growth, one universal base.

In words: keep the exponent, multiply it by the natural log of the base. In letters, for any $a > 0$:

$a^x = e^{x \ln a}$

One line, a rewrite in terms of base e for every base. Check it again: $5^3 = 125$, and $e^{3\ln 5} = 125$. Both say 125.

This identity is the entire skill. Hall and Knight's classic derives it as the Exponential Theorem: every base-$a$ power is secretly an $e$ power. That is what a rewrite in terms of base e reveals. When a problem says rewrite in terms of base e, this is the tool.

Every rewrite in terms of base e runs on three moves. Granville's classic calculus states the engine: "take the logarithm of both sides to the base e."

Step 1 - take ln of both sides. For $y = a^x$, write $\ln y = \ln(a^x)$.

Step 2 - pull the exponent down. The power rule $\ln(a^x) = x\ln a$ turns it into $\ln y = x\ln a$.

Step 3 - undo the ln. Raise $e$ to both sides: $y = e^{x\ln a}$. Done - the exponential side of a rewrite in terms of base e.

Logs convert too. Matthews' 1890 manual proves $\log_b a \cdot \log_c b = \log_c a$, which rearranges into the change-of-base rule $\log_b x = \dfrac{\ln x}{\ln b}$ - the logarithm's rewrite in terms of base e.

Mini drills: $2^x$ becomes $e^{x\ln 2}$, and $\log_4 64$ becomes $\tfrac{\ln 64}{\ln 4}$ - each a one-line rewrite in terms of base e. Your calculator ships only log and ln keys: to rewrite in terms of base e is a daily tool, not a party trick.

Problem. Do a rewrite in terms of base e for each expression: (a) $5^x$ (b) $(\tfrac{1}{2})^h$.

Solution (a). Apply the identity to rewrite in terms of base e: $5^x = e^{x\ln 5}$. Since $\ln 5 \approx 1.6094$, that is $e^{1.6094x}$.

Check at $x = 3$: $e^{3\ln 5} = 125 = 5^3$. The rewrite in terms of base e lost nothing.

Solution (b). $(\tfrac{1}{2})^h = e^{h\ln(1/2)}$. Now $\ln(1/2) = -\ln 2 \approx -0.6931$, so $(\tfrac{1}{2})^h = e^{-h\ln 2}$.

Answer. (a) $e^{x\ln 5}$; (b) $e^{-h\ln 2}$.

Bases below 1 pick up negative exponents: decay is still growth after a rewrite in terms of base e - just negative growth.

Problem. Your calculator has no $\log_4$ key. Use ln to evaluate $\log_4 64$.

Solution. Change of base turns it into natural logs: $\log_4 64 = \tfrac{\ln 64}{\ln 4}$.

Numerator: $\ln 64 \approx 4.1589$. Denominator: $\ln 4 \approx 1.3863$.

Divide: $\tfrac{4.1589}{1.3863} = 3.000$. Verify by definition: $4^3 = 64$. True.

Answer. $3$.

This is the log half of a rewrite in terms of base e: any log, any base, becomes one ln over another. This log-side rewrite in terms of base e recurs constantly from here on.

Problem. Deposit 1000 dollars at 5% annual interest for 10 years. Find the final amount (a) compounded annually, (b) compounded continuously. Round to cents.

Solution (a). Annual compounding stays in its own base - no rewrite in terms of base e needed: $1000 \cdot (1.05)^{10} = 1628.89$.

Solution (b). Continuous compounding uses the ceiling formula $A = Pe^{rt}$ - the continuous rewrite in terms of base e. With $P = 1000$, $r = 0.05$, $t = 10$: $A = 1000 \cdot e^{0.5} = 1648.72$.

Continuous beats annual by 19.83 dollars here. Small gap, cleaner formula - finance chose to rewrite in terms of base e long ago.

Answer. (a) 1628.89 dollars; (b) 1648.72 dollars.

The opening scene closes: push compounding to every moment and $(1+\tfrac{r}{q})^{qn}$ becomes $Pe^{rn}$ - Hall and Knight's 1887 derivation, the origin story of every rewrite in terms of base e.

1. Writing "in" instead of ln. ln abbreviates logarithmus naturalis - no letter i. Pronounce it "ell-enn"; typing in 5 fails, and your rewrite in terms of base e dies before it starts.

2. Forgetting to pull the exponent down. Correct: $\ln(5^x) = x\ln 5$. Wrong: $(\ln 5)^x$. The exponent multiplies the whole log in a rewrite in terms of base e.

3. Dropping the parentheses in $e^{x\ln a}$. All of $x\ln a$ is the exponent. Reading it as $e \cdot x \cdot \ln a$ fails a one-second check: try $x = 1$, $a = e$.

4. Flipping the change-of-base fraction. Correct: $\log_4 64 = \tfrac{\ln 64}{\ln 4} = 3$. Flipped: $\tfrac{\ln 4}{\ln 64} = \tfrac{1}{3}$ - the reciprocal, not the answer.

5. Expecting compound interest to blow up. However large $n$ gets, $(1+\tfrac{1}{n})^n$ stays under $e \approx 2.71828$. Never swap formulas either: continuous means $e^{rt}$, annual means $(1+r)^t$ - the two cases of a rewrite in terms of base e students mix most.