Grade 11

Exponential and logarithmic functions

Functions modeling repeated multiplication and its inverse, used for growth, decay and scales like pH or decibels.

IntuitionGrowth That Feeds on Itself

A bacteria colony that doubles every hour, money earning compound interest, and a radioactive sample losing a fixed fraction each year all share the same shape: the amount already present determines how much gets added next, so the total is multiplied by a constant factor $a$ at every step — giving the exponential function $a^x$ . The logarithmic function $\log_a x$ answers the reverse question: "how many steps of multiplying by $a$ does it take to reach $x$ ?" — exactly the tool needed to solve for time in a growth problem, or to compress a huge range of values (like sound intensity or acidity) into a manageable scale.

The degree-

3

Taylor approximation

1 + x + \frac{1}{2}x^2 + \frac{1}{5}x^3

of the exponential curve: for

x > 0

each positive derivative bends the graph steeper and steeper upward.

SchoolDefinitions and Basic Laws

Definition: Exponential and logarithmic functions

For a fixed base $a>0$ , $a\ne 1$ , the exponential function $a^x$ is defined for every real $x$ . Its inverse function is the logarithm $\log_a x$ defined for $x>0$ as "the exponent you must raise $a$ to, to get $x$ ": by definition, $y=\log_a x$ means exactly $a^y=x$ . Being inverses of each other means $a^{\log_a x}=x,\quad \log_a(a^x)=x$

a^{x+y} = a^x \cdot a^y

This exponent law — multiplying outputs corresponds to adding exponents — is the algebraic seed from which every logarithm law grows, since a logarithm is defined as an exponent. In particular it will drive the proof, below, that $\log_a(xy) = \log_a x + \log_a y$ for $a>0,\ a\ne 1,\ x,y>0$ .

\log_a(xy) = \log_a x + \log_a y

Basic logarithm laws
Law	Statement (for valid $a,x,y$ )
Product	$\log_a(xy) = \log_a x + \log_a y$
Quotient	$\log_a(x/y) = \log_a x - \log_a y$
Power	$\log_a(x^k) = k\log_a x$
Change of base	$\log_b x = \dfrac{\log_c x}{\log_c b}$

UndergraduateTwo Core Theorems

Theorem: The product law of logarithms

For $a>0,\ a\ne 1,\ x,y>0$ , $\log_a(xy) = \log_a x + \log_a y$ .

Why is it true?

A logarithm is just an exponent in disguise, and exponents add when you multiply the corresponding powers — so multiplication inside a logarithm should turn into addition outside it.

Proof

Let $u=\log_a x$ and $v=\log_a y$ . By the definition of logarithm (it is the inverse of the exponential), this means exactly $a^u=x$ and $a^v=y$ .

Multiply these two equations: $xy = a^u \cdot a^v$ . By the exponent law $a^{u+v}=a^u\cdot a^v$ , the right-hand side equals $a^{u+v}$ , so $xy=a^{u+v}$ .

By definition of logarithm again, $xy=a^{u+v}$ means exactly $\log_a(xy)=u+v$ — but this uses that an exponent producing a given output is unique, i.e. $a^u=a^v \Rightarrow u=v$ whenever $a>0,a\ne 1$ (the exponential function never gives the same output for two different exponents, since it is strictly increasing for $a>1$ or strictly decreasing for $0<a<1$ ).

Substituting back $u=\log_a x$ and $v=\log_a y$ gives $\log_a(xy) = \log_a x + \log_a y$ , which is $\log_a(xy) = \log_a x + \log_a y$ .

Theorem: The change-of-base formula

For valid bases $b,c>0$ , $b,c\ne 1$ , and $x>0$ , $\log_b x = \dfrac{\log_c x}{\log_c b}$ .

Why is it true?

Calculators and computers only implement a logarithm in one or two fixed bases; this formula lets any base be computed from those, by expressing "how many times to multiply by $b$ " in terms of "how many times to multiply by $c$ ".

Proof

Let $y=\log_b x$ . By definition of logarithm, this means $b^y=x$ .

By definition of logarithm again (applied to base $c$ this time), $b$ itself can be written as $b=c^{\log_c b}$ . Substitute this into $b^y=x$ : $(c^{\log_c b})^y = x \;\Longrightarrow\; c^{y\log_c b} = x,$ using the power-of-a-power rule $(c^m)^y=c^{my}$ .

Take $\log_c$ of both sides. By definition, $c^{y\log_c b}=x$ means exactly $\log_c x = y\log_c b$ (the same injectivity fact $a^u=a^v \Rightarrow u=v$ used in the product-law proof guarantees this step is valid, not just "plausible").

Since $c\ne 1$ we have $\log_c b \ne 0$ (because $c^0=1\ne b$ ), so we may divide both sides by $\log_c b$ : $y = \dfrac{\log_c x}{\log_c b}.$ Recalling $y=\log_b x$ gives exactly $\log_b x = \dfrac{\log_c x}{\log_c b}$ .

UndergraduateReal-World Applications and Worked Examples

Exponential functions govern compound interest in finance, population and epidemic growth in biology, and radioactive decay in physics, while logarithms are what let scientists and engineers compress enormous ranges into usable scales: the pH scale for acidity, the decibel scale for sound intensity, and the Richter-type scale for earthquake magnitude are all logarithmic. Solving "how long until a quantity doubles?" or "what input produced this reading?" always means taking a logarithm.

Example: Compound interest and doubling time

An amount $P=1000$ USD is invested at annual interest rate $r=5\%$ , growing as $A = P(1+r)^t$ . How much is it worth after $t=10$ years, and how many years does it take to double (use $\ln 2\approx 0.6931$ , $\ln 1.05\approx 0.04879$ )?

Solution

Substitute $P=1000$ , $r=0.05$ , $t=10$ into $A = P(1+r)^t$ : $A=1000(1.05)^{10}\approx 1000\times 1.6289=1628.9$ , so the investment is worth about $1628.90$ after 10 years.

To find the doubling time we need $A=2P$ , i.e. $2P=P(1+r)^t$ , so $(1+r)^t=2$ . This is exactly a request to solve for an unknown exponent — the defining job of a logarithm.

Take $\ln$ of both sides (any base works, by the change-of-base formula $\log_b x = \dfrac{\log_c x}{\log_c b}$ , so we pick the natural log for convenience): $t\ln(1+r)=\ln 2$ , using the power law of logarithms. Solving, $t=\dfrac{\ln 2}{\ln(1+r)}$ .

Substituting the given values: $t=\dfrac{0.6931}{0.04879}\approx 14.2$ years — a concrete illustration of the "rule of 70/72" approximation used in finance, derived here directly from the logarithm laws rather than assumed.

Example: Computing the pH of a solution

Chemists define acidity by $\text{pH}=-\log_{10}[\mathrm{H^+}]$ , where $[\mathrm{H^+}]$ is the hydrogen ion concentration in mol/L. A vinegar sample has $[\mathrm{H^+}]=10^{-3}$ mol/L. Find its pH, and explain why a pH change of $1$ unit corresponds to a $10\times$ change in concentration.

Solution

Substitute the concentration into the definition: $\text{pH}=-\log_{10}(10^{-3})$ .

By definition of logarithm, $\log_{10}(10^{-3})$ is exactly the exponent to which $10$ is raised to get $10^{-3}$ , which is $-3$ (this is the simplest case of the inverse relationship $a^{\log_a x}=x,\quad \log_a(a^x)=x$ with $a=10$ ). So $\text{pH}=-(-3)=3$ .

For the second part, suppose the concentration changes from $[\mathrm{H^+}]$ to $10\times[\mathrm{H^+}]$ . Using the product law $\log_a(xy) = \log_a x + \log_a y$ with $x=[\mathrm{H^+}]$ , $y=10$ : $\log_{10}(10\cdot[\mathrm{H^+}]) = \log_{10}[\mathrm{H^+}] + \log_{10}10 = \log_{10}[\mathrm{H^+}] + 1$ .

So multiplying the concentration by $10$ adds exactly $1$ to $\log_{10}[\mathrm{H^+}]$ , and therefore subtracts exactly $1$ from the pH (since $\text{pH}=-\log_{10}[\mathrm{H^+}]$ ) — this is precisely why the pH scale is logarithmic: each whole-unit step represents a tenfold change in acidity, which the product law makes precise rather than a rule of thumb.

What is $\log_2 8$ ?

Using the product law $\log_a(xy) = \log_a x + \log_a y$ , simplify $\log_3 4 + \log_3 9$ .

Using the change-of-base formula $\log_b x = \dfrac{\log_c x}{\log_c b}$ , which expression equals $\log_2 5$ ?

A solution has $[\mathrm{H^+}]=10^{-3}$ mol/L. Using $\text{pH}=-\log_{10}[\mathrm{H^+}]$ , what is its pH?