Advanced Algebra

Why they hold. $\log_b x$ is the exponent that turns $b$ into $x$. Multiplying powers of $b$ adds their exponents, so logs turn products into sums; raising a power to the $n$th multiplies its exponent, so logs pull exponents out front. Every log law is an exponent law read backwards. The same view explains the domain: $b^y$ is always positive, so only positive numbers have logarithms.

Spotting it. A sum of logs in one base is the log of a product; after that the equation is a polynomial.

Step 1. $\log_2\big(x(x - 2)\big) = 3$, so $x(x - 2) = 8$, i.e. $x^2 - 2x - 8 = (x - 4)(x + 2) = 0$.

Step 2. The original equation needs $x > 0$ AND $x - 2 > 0$, so $x = -2$ is not allowed. The only solution is $x = 4$.

Check: $\log_2 4 + \log_2 2 = 2 + 1 = 3$ ✓

Spotting it. Different bases, so no exponent matching: take a log of both sides to bring the $x$'s down.

Step 1. $x\ln2 = (x - 1)\ln3$, so $x(\ln3 - \ln2) = \ln3$.

Step 2. $x = \dfrac{\ln3}{\ln3 - \ln2} = \log_{3/2}3 \approx 2.71$.

Why only one solution. $\left(\tfrac32\right)^x = 3$ after dividing by $2^x$ and multiplying by $3$, and $\left(\tfrac32\right)^x$ is increasing, so it hits $3$ once.

Spotting it. A positive integer $N$ has $\lfloor\log_{10}N\rfloor + 1$ digits, since $10^{k} \le N < 10^{k + 1}$ exactly when $N$ has $k + 1$ digits.

Step 1. $\log_{10}2^{100} = 100\log_{10}2 \approx 30.103$.

Step 2. So $10^{30} < 2^{100} < 10^{31}$, and $2^{100}$ has $31$ digits. The fractional part says more: $2^{100} \approx 10^{0.103} \cdot 10^{30} \approx 1.27 \times 10^{30}$, so its leading digit is $1$.

Why it works. For two numbers it is one square: $(\sqrt a - \sqrt b)^2 \ge 0$ expands to $a + b \ge 2\sqrt{ab}$, with equality exactly when $a = b$. For more numbers, the key fact is the same: if the terms are not all equal, replacing the largest and smallest by their average keeps the sum and increases the product, so the product is largest when all terms are equal. In use, you need either the SUM or the PRODUCT to be constant; the other is then bounded.

Spotting it. One constraint, one quantity to minimise: express the area in one variable, then make the product of the terms constant.

Step 1. With base side $x$ and height $h$: $x^2h = 32$, and the area is $x^2 + 4xh = x^2 + \dfrac{128}{x}$.

Step 2. Split the second term: $x^2 + \dfrac{64}{x} + \dfrac{64}{x} \ge 3\sqrt[3]{x^2 \cdot \tfrac{64}{x} \cdot \tfrac{64}{x}} = 3\sqrt[3]{4096} = 3 \cdot 16 = 48$.

Step 3. Equality when $x^2 = \tfrac{64}{x}$, i.e. $x = 4$ and $h = 2$: the area is $16 + 32 = 48$ ✓

Spotting it. Each bracket is a sum of two positive terms, and AM–GM on each introduces $\sqrt a$, $\sqrt b$, $\sqrt c$, whose product is known.

Step 1. $1 + a \ge 2\sqrt a$, and similarly for $b$ and $c$.

Step 2. Multiply the three inequalities (all sides positive): $(1 + a)(1 + b)(1 + c) \ge 8\sqrt{abc} = 8$.

Step 3. Equality needs $a = b = c = 1$, which satisfies $abc = 1$. The least value is $8$.

Spotting it. Write $t = \sqrt{x^2 + 2}$: the expression is $\dfrac{t^2 + 1}{t} = t + \dfrac1t$. AM–GM says $t + \tfrac1t \ge 2$ with equality at $t = 1$, but here $t \ge \sqrt2$, so that equality case is impossible.

Step 1. For $t \ge 1$, $t + \tfrac1t$ is increasing: if $1 \le s < t$ then $\left(t + \tfrac1t\right) - \left(s + \tfrac1s\right) = (t - s)\left(1 - \tfrac{1}{st}\right) > 0$.

Step 2. So the minimum is at the smallest allowed $t = \sqrt2$ (that is, $x = 0$): $\sqrt2 + \tfrac{1}{\sqrt2} = \tfrac{3\sqrt2}{2}$.

For $f(x) + 2f\!\left(\tfrac1x\right) = 3x$, substituting $x \to \tfrac1x$ gives a second equation in the same two unknowns $f(x)$ and $f(1/x)$; solving the pair gives $f(x) = \tfrac{2}{x} - x$. That pairing trick — substitute the transformation that swaps the two unknown values — solves most contest functional equations in two lines.

Contest questions rarely want the whole function: they want $f$ at one point, or the sum of $f$ over a symmetric set. Exploiting $f(x) + f(1-x) = c$ turns a hundred-term sum into $50c$, which is the standard AMC 12 use of symmetry and much faster than finding $f$.

Step 1 — put $x = 2$: $f(2) + 2f\!\left(\tfrac12\right) = 6$.

Step 2 — put $x = \tfrac12$: $f\!\left(\tfrac12\right) + 2f(2) = \tfrac32$.

Step 3 — solve the pair: doubling the second gives $2f(1/2) + 4f(2) = 3$; subtracting the first leaves $3f(2) = -3$, so $f(2) = -1$. Check: the second equation gives $f\!\left(\tfrac12\right) = \tfrac32 + 2 = \tfrac72$, and then $f(2) + 2f\!\left(\tfrac12\right) = -1 + 7 = 6$ ✓. In fact $f(x) = \tfrac2x - x$, and $f(2) = 1 - 2 = -1$.

Spotting it. Setting $y = 1$ turns the equation into a recursion that steps $x$ by $1$, and $f(10)$ is ten steps from $f(0)$.

Step 1. $x = y = 0$: $f(0) = 2f(0)$, so $f(0) = 0$.

Step 2. $y = 1$: $f(x + 1) = f(x) + 3 + 2x$. So $f(n) = f(1) + \sum_{k=1}^{n-1}(3 + 2k) = 3 + 3(n - 1) + n(n - 1) = n^2 + 2n$.

Step 3. $f(10) = 100 + 20 = 120$. Check that $f(x) = x^2 + 2x$ works for all reals: $(x + y)^2 + 2(x + y) = (x^2 + 2x) + (y^2 + 2y) + 2xy$ ✓