Module 03: Composition, inverses, exponentials, and logarithms

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: order matters
C. Precise rules and their domains
D. Why the logarithm laws work
E. How to choose the operation
F. Worked examples
G. Independent practice
H. Complete solutions: Module 03
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Module 02; solving equations and exponent arithmetic.

After

Trace nested inputs, determine an inverse with a domain, use logarithms to undo exponentials, and distinguish inverse functions from reciprocals.

B. The central idea: order matters#

Composition sends one function’s output into another function. If $f(u)=\sqrt{u}$ and $g(x)=x-5$ , then

(f\circ g)(x)=f(g(x))=\sqrt{x-5}.

Read the rightmost operation first. The domain requires both that $g(x)$ exists and that its output belongs to the domain of $f$ . By contrast, $(g\circ f)(x)=\sqrt x-5$ . The first needs $x\ge5$ and the second only $x\ge0$ . Function composition is not generally commutative.

An inverse function undoes a function on specified domains. An exponential changes an input additively into an output multiplicatively; its inverse, a logarithm, turns multiplicative change back into additive change. These structures will later explain both the chain rule and exponential growth.

Order matters. With

f(u)=\sqrt u

and

g(x)=x-5

, the composition

f(g(x))=\sqrt{x-5}

(blue) exists only for

x\ge5

, while

g(f(x))=\sqrt x-5

(orange) exists for

x\ge0

. The two orders give different functions with different domains.

C. Precise rules and their domains#

For $b>0$ , $b\ne1$ , the function $b^x$ has domain $\R$ and range $(0,\infty)$ . Its inverse is $\log_b x$ , defined for $x>0$ :

y=\log_b x\quad\Longleftrightarrow\quad b^y=x.

The natural exponential has base $e$ and inverse $\ln$ . Change of base gives $\log_bx=\ln x/\ln b$ for $x>0$ , $b>0$ , $b\ne1$ ; taking natural logs of $b^y=x$ yields $y\ln b=\ln x$ . One characterization of $e$ is $\lim_{n\to\infty}(1+1/n)^n$ ; later calculus gives equivalent characterizations. For positive $u,v$ and real $r$ ,

\ln(uv)=\ln u+\ln v,\quad \ln(u/v)=\ln u-\ln v,\quad \ln(u^r)=r\ln u.

There is no analogous rule $\ln(u+v)=\ln u+\ln v$ . Exponents distribute over products, not sums: $(uv)^r=u^rv^r$ for positive $u,v$ , whereas $(u+v)^2$ has a middle term.

A function has an inverse on its image exactly when it is one-to-one: different inputs have different outputs. The inverse notation $f^{-1}$ does not mean $1/f$ . The identities are $f^{-1}(f(x))=x$ for inputs of $f$ , and $f(f^{-1}(y))=y$ for outputs in its range. Domain and range exchange roles. These definitions follow the standard function and logarithm treatment [O1F].

D. Why the logarithm laws work#

Write $u=b^s$ and $v=b^t$ , which is possible because $u,v>0$ . Then $uv=b^{s+t}$ . Applying the inverse $\log_b$ gives $\log_b(uv)=s+t=\log_b u+\log_b v$ . The law is not arbitrary symbol manipulation: multiplication of outputs corresponds to addition of exponents. Similarly, $e^{\ln x}=x$ needs $x>0$ , while $\ln(e^x)=x$ is valid for every real $x$ .

A parabola $x^2$ on all of $\R$ has no inverse function because $2$ and $-2$ share output $4$ . Restricting to $x\ge0$ makes it one-to-one and gives inverse $\sqrt x$ . Restricting to $x\le0$ instead gives inverse $-\sqrt x$ . The original domain determines the branch.

E. How to choose the operation#

For a composition, replace the input of the outer function with the entire inner expression. For an inverse, first check one-to-one behavior or a stated restriction, write $y=f(x)$ , solve for $x$ , then state the inverse’s domain. For exponential equations, isolate the exponential before taking logs. For log equations, record positivity before combining logs; algebraic candidates outside that domain are invalid.

F. Worked examples#

Example 1: inverse with a branch. Let $f(x)=(x-2)^2+1$ for $x\ge2$ . From $y-1=(x-2)^2$ , the domain gives $x-2\ge0$ , hence $x=2+\sqrt{y-1}$ . Therefore

f^{-1}(x)=2+\sqrt{x-1},\quad x\ge1.

The reciprocal, $1/[(x-2)^2+1]$ , is entirely different.

Example 2: solve an exponential. The equation $5e^{2t}=17$ gives $e^{2t}=17/5$ . Taking $\ln$ gives $2t=\ln(17/5)$ , and $t=\frac12\ln(17/5)$ . This is exact. A decimal is an approximation, so use $\approx$ rather than $=$ for a rounded value.

Finish the restriction. Solving $\ln(x-1)+\ln(x+1)=\ln8$ suggests $x^2-1=8$ , so $x=\pm3$ . Which survives? Both log inputs must be positive, requiring $x>1$ ; only $3$ survives. This is a domain issue, not a rounding issue.

From the engine: Fate EngineTwo functions that undo each other

A tap lands on the screen in pixels, but the game reasons in world units, so Fate Engine’s 2D camera needs one function from screen to world and another from world to screen. On each axis both are linear: $w = p + n\cdot\text{half}$ with $n = 2s/W - 1$ . A linear function with nonzero slope is one-to-one, so it has an inverse, and the second function is that inverse written out.

Fate Engine: the 2D cameraC++

Vec2 Camera::screenToWorld(const Vec2& screen, int windowWidth, int windowHeight) const {
    // Normalize screen coords to -1..1
    float nx = (screen.x / windowWidth) * 2.0f - 1.0f;
    float ny = 1.0f - (screen.y / windowHeight) * 2.0f; // flip Y

    float halfW = (virtualWidth_ * 0.5f) / zoom_;
    float halfH = (VIRTUAL_HEIGHT * 0.5f) / zoom_;

    return {
        position_.x + nx * halfW,
        position_.y + ny * halfH
    };
}

Vec2 Camera::worldToScreen(const Vec2& world, int windowWidth, int windowHeight) const {
    float halfW = (virtualWidth_ * 0.5f) / zoom_;
    float halfH = (VIRTUAL_HEIGHT * 0.5f) / zoom_;

    float nx = (world.x - position_.x) / halfW;
    float ny = (world.y - position_.y) / halfH;

    return {
        (nx + 1.0f) * 0.5f * windowWidth,
        (1.0f - ny) * 0.5f * windowHeight
    };
}

Check one round trip by hand. With a window 800 pixels wide, a virtual width of 480 world units, zoom 2, and the camera at $x = 100$ , the half-width is $240/2 = 120$ , so 240 units are on screen. A tap at pixel 600 gives $n = 0.5$ and world $x = 160$ . Feeding 160 back gives $n = (160 - 100)/120 = 0.5$ and pixel $(0.5 + 1)(0.5)(800) = 600$ . The domain condition from this module appears as a requirement on the inputs: the zoom, the window size, and the virtual width must not be zero. The $y$ lines differ by one sign because screen $y$ grows downward.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M03.01FSolution

For $f(x)=2x+1$ and $g(x)=x^2-4$ , compute $f(g(x))$ and $g(f(x))$ .

M03.02CSolution

Find the domain of $\ln(\sqrt{7-x})$ .

M03.03CSolution

Find the inverse of $f(x)=3x-7$ and verify both compositions.

M03.04CSolution

Find the inverse of $q(x)=x^2+4$ with domain $x\le0$ . State the inverse’s domain and range.

M03.05CSolution

Solve $4\cdot3^{2x-1}=20$ exactly.

M03.06TSolution

Solve $\ln(x-2)+\ln x=\ln15$ with all domain checks.

M03.07TSolution

A stored quantity follows $Q(t)=80e^{-0.3t}$ for $t\ge0$ , with $t$ in seconds. Find the time when it first reaches $20$ , and interpret the units of $0.3$ .

M03.08SSolution

For $x>0$ , simplify $\ln(x^2)+\ln(1/x)$ in two ways. Explain why the domain would change if one started only with $\ln(x^2)$ .

Optional small hints

For M03.04, $x\le0$ selects the negative square root. For M03.06, positivity is stronger than the domain of the combined polynomial alone.

H. Complete solutions: Module 03#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M03.01Show solution

Order of composition

For $f\circ g$ , substitute $x^2-4$ into $2x+1$ : $2(x^2-4)+1=\boxed{2x^2-7}$ . For $g\circ f$ , substitute $2x+1$ into $x^2-4$ : $(2x+1)^2-4=\boxed{4x^2+4x-3}$ . Both accept every real input, but their formulas differ. At $x=0$ their outputs are $-7$ and $-3$ .

Back to M03.01

M03.02Show solution

The outer function adds a restriction

The square root needs $7-x\ge0$ , but the logarithm needs the square root to be strictly positive. Thus $7-x>0$ , giving $\boxed{(-\infty,7)}$ . At $x=7$ the inner square root exists but equals zero, which cannot be a real logarithm input.

Back to M03.02

M03.03Show solution

Undo a linear rule

Solve $y=3x-7$ for $x$ : $x=(y+7)/3$ . Therefore $f^{-1}(x)=(x+7)/3$ , with domain and range $\R$ . Check $f^{-1}(f(x))=(3x-7+7)/3=x$ , and $f(f^{-1}(x))=3(x+7)/3-7=x$ . Both identities hold on the stated domains.

Back to M03.03

M03.04Show solution

The domain selects the root

From $y=x^2+4$ , obtain $x=\pm\sqrt{y-4}$ . The original $x\le0$ selects $x=-\sqrt{y-4}$ . Thus $\boxed{q^{-1}(x)=-\sqrt{x-4}}$ , domain $[4,\infty)$ , range $(-\infty,0]$ . In $q^{-1}(q(x))=-\sqrt{x^2}=-|x|=x$ , the last equality specifically uses $x\le0$ .

Back to M03.04

M03.05Show solution

Isolate before taking logarithms

Divide by $4$ : $3^{2x-1}=5$ . Taking natural logs gives $(2x-1)\ln3=\ln5$ . Since $\ln3\ne0$ ,

\boxed{x=\frac12\left(1+\frac{\ln5}{\ln3}\right)}.

Substitution returns $3^{\ln5/\ln3}=5$ , so the original left side is $20$ .

Back to M03.05

M03.06Show solution

A logarithmic equation

Require $x>2$ . On that domain, combine logs to get $\ln[x(x-2)]=\ln15$ . Since $\ln$ is one-to-one, $x^2-2x=15$ , or $(x-5)(x+3)=0$ . Only $x=5$ satisfies $x>2$ . Checking gives $\ln3+\ln5=\ln15$ , so $\boxed{x=5}$ .

Back to M03.06

M03.07Show solution

A decay threshold

Set $80e^{-0.3t}=20$ , so $e^{-0.3t}=1/4$ . Taking logs gives $-0.3t=-\ln4$ , hence $\boxed{t=\ln4/0.3\text{ s}}$ , approximately $4.62$ seconds. The exponent must be dimensionless, so the rate constant is $0.3\,\mathrm{s}^{-1}$ . Since the exponential decreases strictly for $t\ge0$ , this threshold time is unique.

Back to M03.07

M03.08Show solution

Equivalent routes need the same domain

On $x>0$ , the power and reciprocal laws give $2\ln x-\ln x=\ln x$ . Alternatively, combine products: $\ln[x^2(1/x)]=\ln x$ . The original combined expression needs $1/x>0$ , hence $x>0$ . In isolation, $\ln(x^2)$ is defined for every $x\ne0$ and equals $2\ln|x|$ . Dropping the absolute value would incorrectly erase negative inputs from that larger domain.

Back to M03.08

I. Retention and readiness#

Mix M02.03 with M03.02, and M01.06 with M03.04. Explain why $f^{-1}$ and $1/f$ perform different jobs. Advance when you can handle a fresh logarithmic equation and explain which domain restriction rejects an algebraic candidate.