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Exponential and logarithm

Level 7 · Ages 17–18

Quick recall· Let's wake up what you already know before we start.

Simplify 2³ × 2⁴.

The whole lesson

Read it again any time, including while you work through the exercises.

A bacteria population doubling every hour, savings earning compound interest, information spreading: all these multiplicative growths explode much faster than any straight line. To describe them in continuous time, mathematicians use the exponential function, written exp(x) or e to the power x, where e is a number close to 2.718.

The exponential extends the power rules you already know. Golden rule: e to the a times e to the b equals e to the (a + b). A product of exponentials translates into a SUM of exponents, not a product! Likewise, (e to the a) raised to the power b gives e to the (a × b), and e to the 0 equals 1. Another key property: e to the x is always strictly positive, whatever the value of x, and the function is strictly increasing.

The natural logarithm, written ln, is the mirror function of the exponential: it answers the reverse question. If e to the x equals 5, then x equals ln(5). The two functions cancel each other: ln(e to the x) = x and e to the ln(x) = x for x strictly positive. Careful: ln only accepts strictly positive numbers, because no exponential ever gives a negative or zero result.

The logarithm inherits the mirrored rules: ln(a × b) = ln(a) + ln(b), the logarithm turns products into sums. That is its historical superpower: before calculators, it turned hard multiplications into easy additions. Also remember ln(1) = 0 and ln(e) = 1.

For equations, the strict growth of both functions gives a crystal clear method: e to the a equals e to the b is equivalent to a = b, and to solve e to the x equals k (with k positive), apply ln to both sides: x = ln(k).

The worked example, step by step

Simplify e² × e⁵, then (e³)², then ln(e⁴).

  1. e² × e⁵ = e²⁺⁵ = e⁷

    A product of exponentials: I add the exponents.

  2. (e³)² = e³ˣ² = e⁶

    A power of a power: I multiply the exponents.

  3. ln(e⁴) = 4

    ln and exp are mirror functions: they cancel out, leaving the exponent.

Result: e² × e⁵ = e⁷, (e³)² = e⁶ and ln(e⁴) = 4.

Solve the equation e^x = 5.

  1. 5 > 0

    I first check that the right side is strictly positive, otherwise no solution.

  2. ln(e^x) = ln(5)

    I apply ln to both sides: the equality is preserved because ln is strictly increasing.

  3. x = ln(5) ≈ 1,61

    ln(e^x) = x, so the exact solution is ln(5). The approximation confirms: e^1.61 ≈ 5.

Result: The solution is x = ln(5).