Functions, Exponents & Logarithms

1.Composition and inverses

A function is a rule that assigns each input exactly one output, and composition chains two rules: $(f\circ g)(x)=f(g(x))$. The first thing to notice is that order matters. Applying $g$ first and then $f$ is almost never the same as applying them in the other order. With $f(x)=x+3$ and $g(x)=x^2$, we get $f(g(x))=x^2+3$ but $g(f(x))=(x+3)^2$, and these already disagree at $x=0$.

To evaluate a composition, work from the inside out. Evaluate the innermost function at the given input, then feed that number into the next function. Many contest errors come from computing $f(2)$ when the problem asks for $f(g(2))$. A clean habit is to write the intermediate value on scratch paper, such as $g(2)=5$, before applying $f$ to it.

A function $f$ has an inverse exactly when it is one-to-one on its domain, meaning distinct inputs always give distinct outputs (the horizontal line test). The defining property is $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$. In practice you rarely need a formula for $f^{-1}$. To compute $f^{-1}(c)$, solve $f(x)=c$ for $x$; the solution is $f^{-1}(c)$. This turns an inverse question into an ordinary equation, which is the most useful fact in the lesson.

Why does this work? If $f$ is one-to-one and $f(a)=c$, then $a$ is the only input with output $c$, so sending $c$ back to $a$ is a well-defined rule, and that rule is $f^{-1}$. Two consequences follow. First, $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$: the order reverses, like taking off socks after shoes. Second, if $f(x)=ax+b$ with $a\neq 0$, then $f^{-1}(y)=\frac{y-b}{a}$, which you can confirm by composing: $f\!\left(\frac{y-b}{a}\right)=y$.

Injectivity is the hidden hypothesis. The map $x\mapsto x^2$ has no inverse on all of $\mathbb{R}$ because $2$ and $-2$ both map to $4$. A composition can therefore fail to be invertible even when both pieces are nice, and you must check one-to-one-ness on the domain you are actually using before writing $f^{-1}$ or $(f\circ g)^{-1}$.

• $(f\circ g)(x)=f(g(x))$: evaluate the inner function first
• $f^{-1}(c)=a$ exactly when $f(a)=c$, provided $f$ is one-to-one
• $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$
• If $f(x)=ax+b$, then $f^{-1}(y)=\dfrac{y-b}{a}$

• Evaluating the outer function first, so $f(g(2))$ is treated as $g(f(2))$
• Reporting $f^{-1}(c)$ as $1/f(c)$ instead of the input that produces $c$

2.Iteration, fixed points and periodicity

Iterating a function means applying it repeatedly: $f^{2}(x)=f(f(x))$, $f^{3}(x)=f(f(f(x)))$, and in general $f^{n}$ is the $n$-fold composition. This is not the $n$th power of $f(x)$; the superscript counts compositions, and context usually makes this clear. Iteration questions typically ask for $f^{n}(x_0)$ with $n$ huge, and the trick is almost never to carry out $n$ steps.

The key object is the orbit of a starting point: $x_0,\ x_1=f(x_0),\ x_2=f(x_1),\dots$. Compute terms until a value repeats. If $x_j=x_k$ for some $j<k$, then the sequence is periodic from index $j$ on, with period $p=k-j$. Once you know the period, $x_n$ depends only on $n$ modulo $p$, which reduces $f^{2024}$ to a computation with at most $p$ steps.

Why does periodicity follow? Because $f$ is a function, the next term is determined by the current term alone. If $x_j=x_k$, applying the same function to equal inputs gives $x_{j+t}=x_{k+t}$ for every $t\ge 0$, by induction. That single fact is what makes orbits predictable: once a value recurs, the entire future repeats.

A fixed point is a value with $f(a)=a$, and it is a period-$1$ point. A point of period $2$ satisfies $f(f(a))=a$ but $f(a)\neq a$. So the equation $f^{2}(x)=x$ always contains the fixed points of $f$, and the genuinely new solutions are the period-$2$ points. Factoring out the fixed-point solutions is the standard way to count solutions of $f^{k}(x)=x$.

Not every orbit repeats. For $f(x)=2x$ we have $f^{n}(x)=2^{n}x$, which grows without bound, and such questions are about growth and exponent laws instead. Whenever a period $p$ exists, $f^{p}$ fixes every point of the orbit, so $f^{n}=f^{\,n \bmod p}$ on that orbit. This is the algebraic content of the modular reduction, and it is the form in which the idea is used on the exam.

• $f^{n}$ denotes the $n$-fold composition of $f$, not the $n$th power of $f(x)$
• If $x_j=x_k$ with $j<k$, the orbit is periodic with period $k-j$
• Fixed points satisfy $f(a)=a$, and they also satisfy $f^{2}(a)=a$
• On a $p$-cycle, $f^{n}(x_0)=f^{\,r}(x_0)$ where $r\equiv n \pmod p$

• Reading $f^{n}(x)$ as $(f(x))^{n}$ and then computing a power instead of iterating
• Stopping the orbit too early and assuming a cycle has length $2$ when it is longer

3.Functional equations: special values and symmetry

A functional equation determines a function from a relation it must satisfy for every input, such as $f(x+y)=f(x)+f(y)$. Unlike a single equation in one unknown, it gives infinitely many constraints at once, and the strategy is to choose inputs that make the relation simple. The first step is almost always to substitute special values such as $x=0$, $x=1$, $x=-1$, or a reciprocal.

Special values often reveal structure immediately. Setting $x=y=0$ in $f(x+y)=f(x)+f(y)$ gives $f(0)=2f(0)$, so $f(0)=0$. If the relation mentions both $f(x)$ and $f(1/x)$, then substituting $1/x$ for $x$ produces a second equation in the same two unknowns. Two equations in two unknowns can be solved by elimination, exactly as in linear algebra.

The symmetry trick is the central idea. Suppose $f(x)+2f\!\left(\frac1x\right)=3x$ for all $x\neq 0$. Replacing $x$ by $\frac1x$ gives $f\!\left(\frac1x\right)+2f(x)=\frac3x$. Multiply this second equation by $2$ and subtract the first: $3f(x)=\frac6x-3x$, so $f(x)=\frac2x-x$. Substituting back checks the answer, and the check is not optional, because some functional equations have no solution, and an inconsistent system shows up as a contradiction.

Why does elimination work? Each substitution yields a valid identity, because the relation holds at every input. Following the substitution map through its orbit (for $x\mapsto \frac1x$ the orbit has length $2$, and for $x\mapsto\frac{1}{1-x}$ it has length $3$) produces one linear equation per point of the orbit. The values of $f$ on the orbit are the unknowns, and the determinant of the system decides whether the answer is unique.

Domains need the same care as in any algebra problem. The relation $f(x)+2f(1/x)=3x$ only makes sense for $x\neq 0$, so a substitution that sends $x$ to $0$ is forbidden. Likewise, substituting $1-x$ is valid only if the relation holds at $1-x$. Record the domain of every substitution, because an answer derived from an invalid substitution is not an answer.

• Setting $x=y=0$ in $f(x+y)=f(x)+f(y)$ forces $f(0)=0$
• If $f(x)+2f(1/x)=g(x)$ for all $x\neq 0$, substituting $1/x$ gives a second equation
• Eliminate the orbit values to solve the resulting linear system
• Always check the candidate function in the original relation

• Substituting a value outside the domain on which the relation holds
• Solving the system but never checking the answer in the original equation

4.Exponent laws

The exponent laws are $a^{m}a^{n}=a^{m+n}$, $(a^{m})^{n}=a^{mn}$, $(ab)^{n}=a^{n}b^{n}$, $a^{-n}=\frac1{a^{n}}$, and $\frac{a^{m}}{a^{n}}=a^{m-n}$. They look like rules to memorize, but they share one principle: exponentiation turns addition in the exponent into multiplication. Contest problems are almost always solved by rewriting every term as a power of one common base.

The product law is easy to justify for positive integers: $a^{m}a^{n}$ is $m$ copies of $a$ times $n$ further copies, which is $m+n$ copies. Extending to all integers requires defining $a^{0}=1$ and $a^{-n}=\frac1{a^n}$ so that the laws remain consistent. For rational exponents, $a^{p/q}=\sqrt[q]{a^{p}}$, and the laws continue to hold. Real exponents are defined by continuity, and the exam rarely needs more than rationals.

The most useful technique is to factor out the smallest power. For instance $2^{n+3}-2^{n+1}=2^{n}\left(2^{3}-2^{1}\right)=6\cdot 2^{n}$. Similarly $9^{n}+3^{2n+1}$ becomes $3^{2n}+3\cdot 3^{2n}=4\cdot 3^{2n}$, since $9^{n}=3^{2n}$. Whenever a sum of powers appears, look for a common base and then for a common factor that can be pulled out.

Mixed bases are the usual obstacle. The numbers $4=2^{2}$, $8=2^{3}$, $9=3^{2}$, $16=2^{4}$ and $27=3^{3}$ should be recognized immediately. Then $\frac{8^{x}}{4^{x}}=\frac{2^{3x}}{2^{2x}}=2^{x}$, a simplification that carries straight into exponential equations. Noticing that two bases share a prime is the most frequent first step in these problems.

The common errors come from distributing exponents over sums, which is false: $(a+b)^{n}\neq a^{n}+b^{n}$. Also $2^{n}+2^{n}=2^{n+1}$, not $4^{n}$. The rule $(a^{m})^{n}=a^{mn}$ does not give $a^{m^{n}}$, and $(-1)^{k}$ depends on the parity of $k$, so keep track of signs when exponents are variable.

• $a^{m}a^{n}=a^{m+n}$, $\ (a^{m})^{n}=a^{mn}$, $\ a^{-n}=\dfrac{1}{a^{n}}$
• $\dfrac{a^{m}}{a^{n}}=a^{m-n}$ for $a\neq 0$
• $a^{m}+a^{m}=2a^{m}$, but $a^{m}+a^{m}\neq a^{2m}$
• Rewrite every term as a power of a single base, then factor out the smallest power

• Distributing an exponent over a sum, writing $(a+b)^{n}=a^{n}+b^{n}$
• Adding like powers incorrectly, such as $2^{n}+2^{n}=4^{n}$

5.Solving exponential equations

An exponential equation has the unknown in an exponent. There are two main tools. The first is to equate exponents when both sides are powers of the same base: if $a>0$, $a\neq 1$, and $a^{u}=a^{v}$, then $u=v$. The second is to substitute $t=a^{x}$, which turns the equation into an algebraic one, usually a polynomial in $t$. Contest problems mix bases that share a core, so the second tool is the one used most.

Why can exponents be equated? For $a>0$ with $a\neq 1$, the function $x\mapsto a^{x}$ is strictly monotonic, so it is one-to-one and has an inverse $\log_a$. Applying $\log_a$ to both sides of $a^{u}=a^{v}$ gives $u=v$. Monotonicity also means each piece of an exponential equation has at most one solution in each interval where the expression is defined, which is why sign analysis and substitution are so effective.

For the substitution, write every term in terms of one quantity. With $t=2^{x}$ we have $4^{x}=(2^{x})^{2}=t^{2}$ and $2^{x+1}=2t$. The equation $4^{x}-6\cdot 2^{x}+8=0$ becomes $t^{2}-6t+8=0$, which factors as $(t-2)(t-4)=0$. Each positive root gives one solution $x=\log_{2}t$, namely $x=1$ and $x=2$. Any root with $t\le 0$ must be discarded, since $2^{x}$ is always positive.

Sum questions can often be finished without solving. If $t_1,t_2$ are the roots of the quadratic in $t$, the corresponding exponents are $\log_2 t_1$ and $\log_2 t_2$, and their sum is $\log_2(t_1t_2)$. Here $t_1t_2=8$, so the sum of the $x$-values is $\log_2 8=3$, which agrees with $1+2$. Using the product of the $t$-roots is a reliable shortcut whenever the question asks for a sum of solutions.

Typical errors include forgetting that $2^{x}>0$ eliminates negative roots of the quadratic, and dividing both sides by an expression that might be zero. Also write $4^{x}$ as $(2^{x})^{2}$ rather than treating $2^{2x}$ as a new unrelated variable, since the substitution only works when the whole equation is expressed through one quantity. Verify each candidate in the original equation.

• If $a>0$, $a\neq 1$, then $a^{u}=a^{v}$ implies $u=v$
• Substitute $t=a^{x}$ and write each power as a power of $t$
• Discard roots with $t\le 0$, because $a^{x}>0$
• For a sum of solutions, the sum of the logarithms of the $t$-roots is $\log_a(t_1t_2)$

• Keeping a negative or zero value of $t$, which corresponds to no real $x$
• Treating $4^{x}$ as an independent unknown instead of $(2^{x})^{2}$

6.Logarithm laws and change of base

The logarithm $\log_a b$ is the exponent to which $a$ must be raised to produce $b$, so $a^{\log_a b}=b$. It is the inverse of the exponential, and every logarithm law is an exponent law read in reverse. For $a>0$, $a\neq 1$ and $x,y>0$ we have $\log_a(xy)=\log_a x+\log_a y$, $\log_a\!\left(\frac xy\right)=\log_a x-\log_a y$, and $\log_a(x^{r})=r\log_a x$.

To justify the product law, let $u=\log_a x$ and $v=\log_a y$, so $x=a^{u}$ and $y=a^{v}$. Then $xy=a^{u+v}$, and taking $\log_a$ gives $\log_a(xy)=u+v$. The power law follows the same way, since $x^{r}=a^{ru}$. Each law is simply the corresponding exponent law, which is why it is worth remembering the logic rather than only the formulas.

The change-of-base formula says that for any positive $c\neq 1$, $\log_a b=\frac{\log_c b}{\log_c a}$. To prove it, let $t=\log_a b$, so $a^{t}=b$. Taking $\log_c$ of both sides gives $t\log_c a=\log_c b$, and dividing gives the formula. This identity drives most logarithm simplifications: rewrite everything in one base, often the base that makes the answer clean, and the problem becomes a ratio.

Two consequences are used constantly. First, $\log_a b\cdot\log_b a=1$. Second, the chain $\log_2 3\cdot\log_3 4\cdots\log_{n}(n+1)$ telescopes to $\log_2(n+1)$, because each denominator cancels the next numerator. Any product in which the argument of one logarithm is the base of the next should trigger this telescoping instinct.

Here is a model simplification. Evaluate $\log_4 27\cdot\log_9 16$ by writing both logarithms in base $2$. We have $\log_4 27=\frac{\log_2 27}{\log_2 4}=\frac32\log_2 3$ and $\log_9 16=\frac{\log_2 16}{\log_2 9}=\frac{4}{2\log_2 3}=\frac{2}{\log_2 3}$. Their product is $3$. The expression looked unrelated until the common base was chosen. A poor choice of base leaves irrational logarithms that hide the structure.

• $\log_a(xy)=\log_a x+\log_a y$ and $\log_a(x^{r})=r\log_a x$ for $x,y>0$
• $\log_a b=\dfrac{\log_c b}{\log_c a}$ for any $c>0$, $c\neq 1$
• $\log_a b\cdot\log_b a=1$
• Telescoping: $\log_2 3\cdot\log_3 4\cdots\log_{n}(n+1)=\log_2(n+1)$

• Writing $\log(x+y)=\log x+\log y$, which is false
• Swapping the base and the argument when applying the change-of-base formula

7.Log equations and domain traps

A logarithm is defined only when its argument is positive and its base is positive and different from $1$. When you solve $\log_a f(x)=\log_a g(x)$, the algebra reduces it to $f(x)=g(x)$, and that step is valid only for points where both sides are defined. Every root of the algebraic equation must be checked against $f(x)>0$ and $g(x)>0$. Failing to check is the most common reason a logarithm answer is wrong.

Why does the reduction lose information? The function $u\mapsto\log_a u$ is one-to-one on $(0,\infty)$, so $\log_a f=\log_a g$ implies $f=g$ provided both are positive. The polynomial equation $f=g$, however, has solutions on the whole real line, including points where $f$ or $g$ is negative. Those points are extraneous: the domain is exactly what the derivation silently assumed.

Consider $\log_3(x^{2}-1)=\log_3(2x+2)$. Equating arguments gives $x^{2}-2x-3=0$, so $x=3$ or $x=-1$. For $x=-1$ we get $x^{2}-1=0$, which is not positive, so $x=-1$ is extraneous. For $x=3$ we get $x^{2}-1=8>0$ and $2x+2=8>0$, so $x=3$ is a genuine solution. There is exactly one solution.

A reliable method is to write the domain as inequalities first, solve the equation, and intersect the solutions with the domain. The domain step often removes half the roots of a quadratic. For a sum of logarithms, $\log_a f+\log_a g=\log_a(fg)$ is valid when $f>0$ and $g>0$ individually. The condition $fg>0$ is not enough, because a product of two negatives is positive and would admit spurious solutions.

Equations with the unknown in the base need a different substitution. Consider $\log_2 x+\log_x 2=\frac52$. Let $t=\log_2 x$, which requires $x>0$ and $x\neq 1$. Then $\log_x 2=\frac1t$, and the equation becomes $t+\frac1t=\frac52$, with roots $t=2$ and $t=\frac12$. These give $x=4$ and $x=\sqrt2$, both admissible, and the domain check here is the condition $t\neq 0$.

• $\log_a u$ is defined only for $u>0$, $a>0$, $a\neq 1$
• $\log_a f+\log_a g=\log_a(fg)$ requires $f>0$ and $g>0$ separately
• Intersect the solution set of an algebraic reduction with the domain
• If the unknown is the base, substitute $t=\log_2 x$ and solve for $t$

• Keeping a root that makes an argument zero or negative
• Combining logarithms into $\log_a(fg)$ without checking $f>0$ and $g>0$

8.Floor, ceiling and fractional part

The floor $\lfloor x\rfloor$ is the largest integer not exceeding $x$, the ceiling $\lceil x\rceil$ is the smallest integer not less than $x$, and the fractional part is $\{x\}=x-\lfloor x\rfloor\in[0,1)$. Everything begins from the defining inequality $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$. Equivalently, write $x=n+\{x\}$ with $n$ an integer and $0\le\{x\}<1$, and most floor equations become a problem about intervals.

Useful identities: $\lfloor x+n\rfloor=\lfloor x\rfloor+n$ for every integer $n$, so integer shifts can be pulled out of a floor, but scalar multiples cannot: $\lfloor 2x\rfloor\neq 2\lfloor x\rfloor$ in general. For non-integer $x$, $\lfloor x\rfloor+\lfloor -x\rfloor=-1$, and always $\lceil x\rceil=-\lfloor -x\rfloor$. Negative numbers round down under the floor, so $\lfloor -2.5\rfloor=-3$.

Counting with floors reduces to counting blocks. The value $\lfloor\sqrt n\rfloor=k$ holds exactly when $k^{2}\le n\le (k+1)^{2}-1$, a block of $2k+1$ consecutive integers. Likewise $\lfloor n/m\rfloor=q$ holds exactly for $n$ in $[qm,\,qm+m-1]$, a block of $m$ integers. Summing over blocks replaces a long sum of $N$ terms by a sum of about $\sqrt N$ block values.

The block method proves that $\sum_{n=1}^{N}\lfloor\sqrt n\rfloor=\sum_{k}k\cdot\#\{n:\lfloor\sqrt n\rfloor=k\}$, and the count of each $k$ is the block length. Converting a sum over $n$ into a sum over the values $k$ is one of the most important tricks in the topic, and the same idea evaluates divisor-type sums such as $\sum_{n}\lfloor N/n\rfloor$ and counts such as the number of $n$ with $\lfloor\log_2 n\rfloor=j$.

Typical mistakes: miscounting block lengths by using $k^{2}$ to $k^{2}+k$ instead of $k^{2}$ to $(k+1)^{2}-1$, and forgetting that the floor of a negative number rounds down. Fractional part questions often reduce to an equation like $x+\{x\}=c$, which you solve by splitting $x$ into its integer and fractional parts and testing the allowed integers.

• $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$, and $\{x\}=x-\lfloor x\rfloor$
• $\lfloor x+n\rfloor=\lfloor x\rfloor+n$ for integers $n$, but $\lfloor 2x\rfloor\neq 2\lfloor x\rfloor$ in general
• For non-integer $x$, $\lfloor x\rfloor+\lfloor -x\rfloor=-1$
• $\lfloor\sqrt n\rfloor=k$ exactly when $k^{2}\le n\le (k+1)^{2}-1$

• Treating $\lfloor -2.5\rfloor$ as $-2$
• Using the wrong block endpoint, such as ending the block for $\lfloor\sqrt n\rfloor=k$ at $k^{2}+k$

9.Comparing huge numbers

Contest problems often ask which of several enormous numbers is largest, or how many digits a power has. Direct computation is impossible, so the method is to compare logarithms, or to rewrite both quantities with a common exponent or base. Taking $\log$ is legitimate because $\log$ is strictly increasing on positive numbers: for $A,B>0$, $A<B$ if and only if $\log A<\log B$. Then $\log(a^{b})=b\log a$ turns a power into a product that can be estimated by hand.

Good estimates need a small toolkit: $\log_{10}2\approx 0.30103$, $\log_{10}3\approx 0.47712$, and $\log_2 3\approx 1.58496$. For example $\log_{10}(2^{300})=300\cdot 0.30103\approx 90.31$, while $\log_{10}(3^{200})=200\cdot 0.47712\approx 95.42$. The second is larger by about $10^{5}$. Keep only as many digits as the comparison requires, and rewrite exactly if two estimates agree to that precision.

Digit counts use the same tool. A positive integer $N$ has $\lfloor\log_{10}N\rfloor+1$ digits, because $10^{d-1}\le N<10^{d}$ exactly when $d-1\le \log_{10}N<d$. For $N=3^{1000}$, $1000\log_{10}3\approx 477.12$, so $3^{1000}$ has $478$ digits. The floor and the $+1$ are where most errors occur, so check the boundary cases explicitly.

The common-exponent trick is equally important. Since $a^{mk}=(a^{m})^{k}=(a^{k})^{m}$, two powers can be compared by writing both with the same exponent and comparing the bases. For instance $2^{100}=(2^{10})^{10}=1024^{10}$ and $10^{30}=(10^{3})^{10}=1000^{10}$, so $2^{100}$ is larger. The same idea shows that $n^{k}$ versus $k^{n}$ is decided by the sign of $\ln n/n-\ln k/k$, which depends on the monotonicity of $\ln x/x$.

The function $\frac{\ln x}{x}$ increases on $(0,e]$ and decreases on $[e,\infty)$. This is the key to inequalities such as $n^{2025}$ versus $2025^{n}$: for $n\ge 3$ and $n<2025$ the larger exponent sits on the wrong side of the peak, while for $n\ge 2025$ the inequality reverses. Mistakes to avoid include comparing only leading digits, rounding logarithms too coarsely near a tie, and forgetting to convert between bases.

• For $A,B>0$: $A<B \iff \log A<\log B$
• A positive integer $N$ has $\lfloor\log_{10}N\rfloor+1$ digits
• $\log_{10}2\approx 0.30103$ and $\log_{10}3\approx 0.47712$
• $a^{mk}=(a^{m})^{k}=(a^{k})^{m}$, so rewrite both sides with a common exponent

• Estimating $\log$ values too roughly to decide a close comparison
• Forgetting the $+1$ in the digit count formula

10.Graph transformations

Given the graph of $y=f(x)$, the graph of a transformed function is obtained by a sequence of standard moves. Replacing $f(x)$ by $f(x-h)+k$ translates the graph right by $h$ and up by $k$. Replacing $f(x)$ by $a\,f(x)$ stretches vertically by $|a|$ and reflects across the $x$-axis when $a<0$. Replacing $f(x)$ by $f(bx)$ compresses horizontally by a factor of $|b|$, and $f(-x)$ reflects across the $y$-axis.

Tracking a single point is the most reliable method. A point $(p,q)$ on $y=f(x)$ produces a point on the transformed graph whose input is found by solving the inner argument for $x$. For $y=a\,f(x-h)+k$, the point $(p,q)$ maps to $(p+h,\ aq+k)$. Check this by substitution: setting $x=p+h$ makes $x-h=p$, so the new output is $aq+k$.

Horizontal and vertical operations behave differently, and the signs are where errors happen. A horizontal change acts on the input before $f$ is applied, so $f(x-h)$ moves points right by $h$, the opposite of the sign inside the bracket. A vertical change acts on outputs and moves points in the direction of its sign: $f(x)+k$ moves points up by $k$. Mixing these two rules is the most common graph error.

Transformations compose like functions. The graph of $y=-f(2-x)+3$ is obtained by three steps. Write $2-x=-(x-2)$, so the input map $p\mapsto 2-p$ is a reflection across the line $x=1$. The output map $q\mapsto -q+3$ is a reflection across the $x$-axis followed by a vertical shift. A point $(p,q)$ on $y=f(x)$ therefore goes to $(2-p,\ 3-q)$. Writing the new function explicitly and then tracking one point avoids confusion.

Mistakes to avoid: applying a horizontal shift to the output, reflecting across the $y$-axis when you meant the $x$-axis, and treating a horizontal stretch by $b$ as moving $p$ to $bp$ instead of $p/b$. When in doubt, plug the candidate point into the new formula and confirm that the output matches.

• $y=a\,f(x-h)+k$ maps $(p,q)$ to $(p+h,\ aq+k)$
• $f(-x)$ reflects across the $y$-axis, and $-f(x)$ reflects across the $x$-axis
• $f(bx)$ maps $(p,q)$ to $\left(\dfrac pb,\ q\right)$
• A point is on the new graph exactly when substituting it into the new formula gives a true statement

• Shifting the wrong way when the shift acts on the input
• Applying a horizontal stretch by $b$ as $p\mapsto bp$ instead of $p\mapsto p/b$

Example 1: Composition order matters

Let $f(x)=x+3$ and $g(x)=x^{2}$. What is $g(f(2))-f(g(2))$?

Example 2: Change of base to a single parameter

Let $a=\log_3 2$. Which expression equals $\log_6 16$?

Example 3: Solve a power equation by one base

Solve $\dfrac{8^{x}}{2^{x+1}}=16$ for $x$.