Exponential and Logarithmic Functions — understand them once, use them forever

Exponential and logarithmic functions are often seen by students as "hard" and "abstract". But these are two of the most powerful mathematical tools for describing reality — once you understand them properly, every problem involving them becomes far more natural.

Exponential functions — the model of growth​

An exponential function has the form:

$f(x) = a^x \quad (a > 0, \, a \neq 1)$

When $a > 1$: the function is increasing (growth) When $0 < a < 1$: the function is decreasing (decay)

Why does it matter? Because many real-world phenomena grow multiplicatively — each period multiplies by $a$: populations, bank savings, infection counts during an epidemic.

The number $e$ — the natural base​

$e \approx 2.718$

The function $f(x) = e^x$ is special because its derivative is itself. This is why $e$ appears everywhere in calculus and physics.

Logarithms — asking "what power?"​

$\log_a b = c \iff a^c = b$

Simply put: logarithm is the inverse of exponentiation. Instead of asking "what is $a$ to the power $c$?", a logarithm asks "what power of $a$ gives $b$?".

Two special logarithms​

• $\log x$ (no base written): defaults to $\log_{10}$ — the common logarithm
• $\ln x$: natural logarithm, base $e$ — widely used in calculus

The most important properties​

$\log_a(mn) = \log_a m + \log_a n$

$\log_a\left(\frac{m}{n}\right) = \log_a m - \log_a n$

$\log_a(m^k) = k \cdot \log_a m$

$\log_a b = \frac{\log_c b}{\log_c a} \quad \text{(change of base)}$

Memory tip: Multiply → add, Divide → subtract, Exponent → bring to front. These three properties let you "break down" any complex logarithmic expression.

Exponential and logarithmic equations — solving strategies​

Exponential equations​

Type 1: Rewrite with the same base $4^x = 8 \implies 2^{2x} = 2^3 \implies 2x = 3 \implies x = \frac{3}{2}$

Type 2: Take the logarithm of both sides $3^x = 7 \implies x \ln 3 = \ln 7 \implies x = \frac{\ln 7}{\ln 3}$

Logarithmic equations​

Strategy: Rewrite as $\log_a f(x) = \log_a g(x)$, then conclude $f(x) = g(x)$, always with the condition $f(x) > 0$.

$\log_2(x+3) = \log_2(2x-1) \implies x+3 = 2x-1 \implies x = 4$

Check: $x = 4 > 0$ ✓ (satisfies the logarithm domain condition)

Real-world applications you've already encountered​

When the real world grows multiplicatively, logarithms convert it to additive scale — much easier to analyze.

Most common mistakes​

• Forgetting the domain: $\log_a x$ is only defined when $x > 0$ and $a > 0, a \neq 1$
• Incorrect rule: $\log(m + n) \neq \log m + \log n$ — a very common error
• Missing extraneous solutions: After solving, always substitute back to verify the domain condition

Upload exponential and logarithmic problems to MathPal to see detailed step-by-step solutions — especially useful when the problem requires substitution or combining multiple properties at once.

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