Math Core

Logarithm problems on the AIME look intimidating, with nested logs, mixed bases and systems of three equations, but they almost always collapse to ordinary algebra. The trick is to pick the right substitution: convert everything to one base, then let the logarithms themselves be your variables.

The rules, and why they work

$\log_b x = y$ means exactly $b^y = x$ (with $b > 0$ , $b \ne 1$ , $x > 0$ ). Every log rule is an exponent rule in disguise:

Log rule	Exponent rule behind it
$\log_b(xy) = \log_b x + \log_b y$	$b^m b^n = b^{m+n}$
$\log_b\!\left(\dfrac{x}{y}\right) = \log_b x - \log_b y$	$\dfrac{b^m}{b^n} = b^{m-n}$
$\log_b(x^k) = k\log_b x$	$(b^m)^k = b^{mk}$
$\log_{b^k} x = \dfrac1k\log_b x$	$(b^k)^{m/k} = b^m$

Change of base

For any valid base $c$ ,

\log_b a = \frac{\log_c a}{\log_c b}.

Consequences: $\log_b a = \dfrac{1}{\log_a b}$ , and logs chain like fractions:

\log_a b \cdot \log_b c \cdot \log_c d = \log_a d.

The chain rule is easiest to see with change of base: $\dfrac{\log b}{\log a} \cdot \dfrac{\log c}{\log b} \cdot \dfrac{\log d}{\log c} = \dfrac{\log d}{\log a}$ . Every middle term cancels, so long products of logs telescope.

One more identity comes up often: $a^{\log_b c} = c^{\log_b a}$ . Take $\log_b$ of both sides and both become $\log_b a \cdot \log_b c$ .

Strategy: make the logs the variables

Pick one base (usually the smallest one in sight, like $2$ ) and rewrite every log in that base.
Substitute $u = \log_2 x$ , $v = \log_2 y$ , and so on. Products and powers of $x, y$ become sums and multiples of $u, v$ .
Solve the algebra (often linear or quadratic) and convert back with $x = 2^u$ .

Exponential equations work the same way in reverse: in $4^x - 3 \cdot 2^{x+1} + 8 = 0$ , let $u = 2^x$ , so $4^x = u^2$ and $2^{x+1} = 2u$ .

Common mistake

Check every solution in the original equation. Logs need positive arguments and bases that are positive and not $1$ . Squaring or combining $\log_b x + \log_b(x - 12)$ into $\log_b\big(x(x - 12)\big)$ can create extraneous roots where $x$ or $x - 12$ is negative.

Worked examples

Worked example: Mixed bases

Solve $\log_2 x + \log_4 x + \log_8 x = 11$ .

Since $\log_4 x = \dfrac12\log_2 x$ and $\log_8 x = \dfrac13\log_2 x$ , the equation is $\left(1 + \dfrac12 + \dfrac13\right)\log_2 x = 11$ , or $\dfrac{11}{6}\log_2 x = 11$ . So $\log_2 x = 6$ and $x = 64$ .

Worked example: A hidden quadratic

Find all real $x$ with $4^x - 3 \cdot 2^{x+1} + 8 = 0$ .

Let $u = 2^x > 0$ . Then $u^2 - 6u + 8 = 0$ , so $u = 2$ or $u = 4$ , giving $x = 1$ or $x = 2$ . Both $u$ values are positive, so both solutions are valid.

Worked example: A system

Positive reals $x, y$ satisfy $\log_2 x + \log_4 y = 4$ and $\log_4 x + \log_2 y = 5$ . Find $xy$ .

Let $a = \log_2 x$ and $b = \log_2 y$ . The system is $a + \dfrac b2 = 4$ and $\dfrac a2 + b = 5$ . Adding gives $\dfrac32(a + b) = 9$ , so $a + b = 6$ and $xy = 2^{a+b} = 64$ . (In fact $a = 2$ and $b = 4$ , so $x = 4$ and $y = 16$ .)

You didn't even need $a$ and $b$ separately. Look for the combination the question asks for.

Worked example: Floors of logs

Compute $\displaystyle\sum_{k=1}^{31}\lfloor\log_2 k\rfloor$ .

$\lfloor\log_2 k\rfloor = j$ exactly when $2^j \le k < 2^{j+1}$ , which happens for $2^j$ values of $k$ . From $1$ to $31$ the blocks are complete:

0 \cdot 1 + 1 \cdot 2 + 2 \cdot 4 + 3 \cdot 8 + 4 \cdot 16 = 0 + 2 + 8 + 24 + 64 = 98.

Tip

When a problem gives several log equations with the same three variables, taking $\log$ of everything usually turns it into a linear system in $\log x, \log y, \log z$ . Add all the equations first. The total is often exactly what you need.