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Solving Real Equations: Half-Life & Interest

Now we put it all to work. Take the log to bring an unknown exponent down to earth, solve exponential and logarithmic equations, and run two classic models: radioactive half-life and compound interest.

The key move: take the log of both sides

When the unknown is stuck up in an exponent — as in 3^x = 20 — you cannot reach it by ordinary algebra. The trick for solving exponential equations is to take the log of both sides, then use the power law to slide the exponent down in front. Once the unknown is a plain multiplier instead of an exponent, it is just a linear equation you can finish off.

Solve  3^x = 20

  ln(3^x) = ln(20)        take ln of both sides
  x·ln(3) = ln(20)        power law brings x down
  x = ln(20) / ln(3)      divide by ln(3)
  x = 2.9957 / 1.0986
  x ≈ 2.727

Check:  3^2.727 ≈ 20.  Good.


Solve a log equation  log_2(x) = 5
  rewrite in exponential form:  x = 2^5 = 32.
Log both sides, drop the exponent with the power law, then divide.

Half-life: solving for time in decay

A radioactive sample shrinks by exponential decay. Its half-life is the time for half of it to disappear — a constant, no matter how much you start with. If the half-life is h, then after time t the fraction remaining is (1/2)^(t/h). Finding *how long* until a given fraction is left means solving for t, and t sits in the exponent — so we take the log.

Carbon-14 has half-life h = 5730 years.
A bone retains 30% of its original C-14. How old is it?

  (1/2)^(t/5730) = 0.30
  ln[(1/2)^(t/5730)] = ln(0.30)
  (t/5730)·ln(1/2) = ln(0.30)        power law
  t/5730 = ln(0.30) / ln(0.5)
  t/5730 = (-1.20397) / (-0.69315)
  t/5730 = 1.7370
  t = 1.7370 · 5730
  t ≈ 9953 years
Half-life decay: set fraction = (1/2)^(t/h), log both sides, solve for t.

Compound interest: money that grows on itself

Money in an account undergoes exponential growth through compound interest. With principal P, annual rate r, compounded n times a year for t years, the balance is A = P·(1 + r/n)^(nt). If interest is compounded *continuously* — the limit from the second guide — it becomes the cleaner A = P·e^(rt), driven by e. To find how long until a target balance, again the unknown t is an exponent, so again we take a log.

$1000 at 5% per year, compounded continuously.
How long to double, to reach $2000?

  2000 = 1000·e^(0.05 t)
  2 = e^(0.05 t)            divide by 1000
  ln(2) = 0.05 t           take ln; ln undoes e
  t = ln(2) / 0.05
  t = 0.69315 / 0.05
  t ≈ 13.86 years

Quick sanity check (the "rule of 70"):
  70 / 5 = 14 years.  Close — the rule
  is just ln(2)·100 ≈ 69.3 rounded to 70.
Continuous growth: divide off P, take ln, and ln cancels the e.

Notice the shape of every problem here. A quantity is modeled as a starting value times a base raised to a power that contains time; we set it equal to a target; we take a log to free the exponent; we solve a linear equation. Decay or growth, money or atoms, the method never changes — that is the quiet power of having built the exponential and its inverse the logarithm from the ground up.