Notebooks · posts/doubling-time

Doubling time

Adam Zettel · September 22, 2026 · written with an agent

Something growing 7% a year doubles in $ln 2/ ln 1.07 = 10.24$ years, not $100/7 = 14.3$ : growth compounds. The logarithm is the tool that turns "how long until" into a division, and the rule of 70 is that division with $ln 2 \approx 0.7$ and $ln (1 + r) \approx r$ .

t_{2} = \frac{ln 2}{ln ( 1 + r )} \approx \frac{70}{100 r}

Growth by a fixed fraction each year is $A (t) = A_{0} (1 + r)^{t}$ . Asking when $A (t) = 2 A_{0}$ means solving $(1 + r)^{t} = 2$ for $t$ , and $t$ is in the exponent; the logarithm brings it down:

t ln (1 + r) = ln 2 ⟹ t = \frac{ln 2}{ln ( 1 + r )} .

Any base works, since a ratio of logs does not depend on the base.

For small $r$ , $ln (1 + r) \approx r$ and $ln 2 \approx 0.693$ , so $t \approx 70/ (100 r)$ : the rule of 70. It is 2.4% off at 7% and 0.5% off at 3%, and a test checks both. The figure draws $1.0 7^{t}$ against the straight line $1 + 0.07 t$ , which is what "7% a year" would mean if growth did not compound.

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[1]lines 26–69

RATE = 0.07

#| $t_2 = \frac{\ln 2}{\ln(1 + r)}$: solve $(1+r)^t = 2$ by taking logs.
def doubling_time(r) {
    "Years for growth at rate r per year to double: ln 2 / ln(1 + r)."
    return log(2.0) / log(1.0 + r)
}
#| $70 / (100r)$: the same thing with $\ln 2 \approx 0.7$ and $\ln(1+r) \approx r$.
def rule_of_70(r) {
    return 70.0 / (100.0 * r)
}

t_double = doubling_time(RATE)
show t_double
show rule_of_70(RATE)

# a point every quarter year for twenty years
N_TIMES = 81
ts = linspace(0.0, 20.0, N_TIMES)
figure("Growth at 7% a year, compounding and not", "years", "amount, starting from 1")
plot(ts, exp(log(1.0 + RATE) * ts), "$1.07^t$")
plot(ts, 1.0 + RATE * ts, "1 + 0.07 t")

test "at 7% a year the doubling time is 10.24 years" {
    assert_near(t_double, 10.2448, 1.0e-3)
}
test "after that long the amount is exactly 2" {
    assert_near(pow(1.0 + RATE, t_double), 2.0, 1.0e-12)
}
test "the rule of 70 is 2.4% off at 7% and 0.5% off at 3%" {
    # how far the rule of 70 is from the exact time, as a fraction of it
    rule70_off_7 = abs(rule_of_70(0.07) - doubling_time(0.07)) / doubling_time(0.07)
    rule70_off_3 = abs(rule_of_70(0.03) - doubling_time(0.03)) / doubling_time(0.03)
    assert_near(rule70_off_7, 0.024, 0.0005)
    assert_near(rule70_off_3, 0.005, 0.0005)
}
test "the base of the logarithm does not matter: base 10 gives the same time" {
    log10_2 = log(2.0) / log(10.0)
    log10_r = log(1.0 + RATE) / log(10.0)
    assert_near(log10_2 / log10_r, t_double, 1.0e-12)
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors

RATE = 0.07

#| $t_2 = \frac{\ln 2}{\ln(1 + r)}$: solve $(1+r)^t = 2$ by taking logs.
def doubling_time(r) {
    "Years for growth at rate r per year to double: ln 2 / ln(1 + r)."
    return log(2.0) / log(1.0 + r)
}
#| $70 / (100r)$: the same thing with $\ln 2 \approx 0.7$ and $\ln(1+r) \approx r$.
def rule_of_70(r) {
    return 70.0 / (100.0 * r)
}

t_double = doubling_time(RATE)
show t_double
show rule_of_70(RATE)

# a point every quarter year for twenty years
N_TIMES = 81
ts = linspace(0.0, 20.0, N_TIMES)
figure("Growth at 7% a year, compounding and not", "years", "amount, starting from 1")
plot(ts, exp(log(1.0 + RATE) * ts), "$1.07^t$")
plot(ts, 1.0 + RATE * ts, "1 + 0.07 t")

test "at 7% a year the doubling time is 10.24 years" {
    assert_near(t_double, 10.2448, 1.0e-3)
}
test "after that long the amount is exactly 2" {
    assert_near(pow(1.0 + RATE, t_double), 2.0, 1.0e-12)
}
test "the rule of 70 is 2.4% off at 7% and 0.5% off at 3%" {
    # how far the rule of 70 is from the exact time, as a fraction of it
    rule70_off_7 = abs(rule_of_70(0.07) - doubling_time(0.07)) / doubling_time(0.07)
    rule70_off_3 = abs(rule_of_70(0.03) - doubling_time(0.03)) / doubling_time(0.03)
    assert_near(rule70_off_7, 0.024, 0.0005)
    assert_near(rule70_off_3, 0.005, 0.0005)
}
test "the base of the logarithm does not matter: base 10 gives the same time" {
    log10_2 = log(2.0) / log(10.0)
    log10_r = log(1.0 + RATE) / log(10.0)
    assert_near(log10_2 / log10_r, t_double, 1.0e-12)
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors