Logarithms and exponentials

This chapter provides a comprehensive review of logarithms and exponentials, covering fundamental laws, transformations, and equation-solving techniques. Mastery of these concepts is crucial for advanced mathematical analysis and is frequently assessed in the CMI examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Logarithm laws | | 2 | Change of base | | 3 | Logarithmic equations | | 4 | Exponential equations | | 5 | Logarithmic inequalities |

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We begin with Logarithm laws.

Part 1: Logarithm laws

Logarithm laws

Overview

Logarithm laws convert multiplication into addition, division into subtraction, and powers into coefficients. In school algebra they are used for simplification and solving equations; in CMI-style questions they also appear in disguised form through functional equations such as $f(xy)=f(x)+f(y)$. So this topic is not just about formulas — it is about structure. ---

Learning Objectives

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Basic Definition

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Domain Conditions

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Main Logarithm Laws

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Reverse Reading of the Laws

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Most Important Traps

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Useful Standard Values

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Structure View: Why Logarithm Laws Matter

This is the discrete analogue of logarithm laws and is exactly the kind of structure behind deeper exam questions. ---

Minimal Worked Examples

Example 1 Simplify $\log_2 48-\log_2 3$. Using quotient law, $\qquad \log_2 48-\log_2 3=\log_2\left(\dfrac{48}{3}\right)=\log_2 16=4$ --- Example 2 Solve $\log_3(x-1)+\log_3(x-3)=2$. First combine: $\qquad \log_3\big((x-1)(x-3)\big)=2$ So, $\qquad (x-1)(x-3)=3^2=9$ $\qquad x^2-4x+3=9$ $\qquad x^2-4x-6=0$ $\qquad x=2\pm \sqrt{10}$ Now check domain:

• $x-1>0$

• $x-3>0$

So we need $x>3$, hence only $\qquad x=2+\sqrt{10}$ is valid. ---

CMI Strategy

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Functional Equation Insight

This is exactly why logarithm laws are useful even far beyond standard school simplification. ---

Practice Questions

:::question type="MCQ" question="For which of the following values is the expression $\log_x 9$ defined over the real numbers?" options=["$x=1$","$x=-3$","$x=\dfrac{1}{2}$","$x=0$"] answer="C" hint="Check the conditions on the base." solution="For $\log_x 9$ to be defined over the reals, the base must satisfy $x>0$ and $x\ne 1$. Among the given values, only $x=\dfrac{1}{2}$ satisfies both conditions. Hence the correct answer is $\boxed{C}$." ::: :::question type="NAT" question="Evaluate $\log_2 48-\log_2 3+\log_2 2$." answer="5" hint="Combine the logarithms into a single logarithm." solution="Using logarithm laws, $\qquad \log_2 48-\log_2 3+\log_2 2=\log_2\left(\dfrac{48\cdot 2}{3}\right)$ $\qquad =\log_2 32=5$ Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true for $a>0$, $a\ne 1$, and positive $x,y$?" options=["$\log_a(xy)=\log_a x+\log_a y$","$\log_a(x+y)=\log_a x+\log_a y$","$\log_a\left(\dfrac{x}{y}\right)=\log_a x-\log_a y$","$\log_a(x^r)=r\log_a x$"] answer="A,C,D" hint="Separate true logarithm laws from false look-alikes." solution="1. True. This is the product law.

False. There is no law of the form $\log_a(x+y)=\log_a x+\log_a y$.

True. This is the quotient law.

True for positive $x$. This is the power law.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Suppose a function $f$ on positive integers satisfies $f(xy)=f(x)+f(y)$ for all positive integers $x,y$. If $f(2)=3$ and $f(3)=-1$, find $f(72)$." answer="8" hint="Use prime factorisation of $72$." solution="Prime-factorise $72$: $\qquad 72=2^3\cdot 3^2$ Using $f(xy)=f(x)+f(y)$ repeatedly, we get $\qquad f(72)=f(2^3)+f(3^2)$ Also, $\qquad f(2^3)=3f(2)=3\cdot 3=9$ and $\qquad f(3^2)=2f(3)=2\cdot (-1)=-2$ Therefore, $\qquad f(72)=9+(-2)=7$ Hence the answer is $\boxed{7}$." ::: ---

Summary

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Part 2: Change of base

Change of Base

Overview

The change of base formula is one of the most useful tools in logarithms. It allows us to rewrite a logarithm in a more convenient base, compare logarithmic expressions, simplify algebraic forms, and evaluate logs using known values. In CMI-style algebra, this topic is often tested through identities, cyclic products, simplification, and careful domain checking. ---

Learning Objectives

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Logarithm Basics Needed First

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Main Formula

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Why the Formula Works

Derivation Let $\qquad x=\log_a b$ Then by definition, $\qquad a^x=b$ Taking logarithm in base $c$ on both sides, $\qquad \log_c(a^x)=\log_c b$ Using the power rule, $\qquad x \log_c a = \log_c b$ So, $\qquad x=\dfrac{\log_c b}{\log_c a}$ Hence, $\qquad \log_a b = \dfrac{\log_c b}{\log_c a}$ ---

Most Important Consequences

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Domain Conditions

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Fast Simplification Patterns

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Common Traps

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Minimal Worked Examples

Example 1 Evaluate $\log_4 32$. Using change of base, $\qquad \log_4 32 = \dfrac{\log 32}{\log 4}$ Now write $32=2^5$ and $4=2^2$: $\qquad \dfrac{\log(2^5)}{\log(2^2)}=\dfrac{5\log 2}{2\log 2}=\dfrac{5}{2}$ So, $\qquad \log_4 32=\dfrac{5}{2}$ --- Example 2 Simplify $\log_2 5 \cdot \log_5 7 \cdot \log_7 16$. Using the chain identity, $\qquad \log_2 5 \cdot \log_5 7 = \log_2 7$ So the whole product is $\qquad \log_2 7 \cdot \log_7 16 = \log_2 16 = 4$ Hence the value is $4$. ---

High-Value Algebraic Forms

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is equal to $\log_a b$ whenever both sides are defined?" options=["$\dfrac{\log a}{\log b}$","$\dfrac{\ln b}{\ln a}$","$\log_b a$","$\dfrac{1}{\ln b}$"] answer="B" hint="Recall the change of base formula." solution="By the change of base formula, $\qquad \log_a b = \dfrac{\ln b}{\ln a}$. So the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Evaluate $\log_4 32 + \log_9 27$." answer="4" hint="Rewrite both using powers of $2$ and $3$." solution="We compute each term separately. First, $\qquad \log_4 32 = \dfrac{\log 32}{\log 4} = \dfrac{5\log 2}{2\log 2} = \dfrac{5}{2}$ Next, $\qquad \log_9 27 = \dfrac{\log 27}{\log 9} = \dfrac{3\log 3}{2\log 3} = \dfrac{3}{2}$ Therefore, $\qquad \log_4 32 + \log_9 27 = \dfrac{5}{2} + \dfrac{3}{2} = 4$ Hence the answer is $\boxed{4}$." ::: :::question type="MSQ" question="Which of the following statements are true whenever all expressions involved are defined?" options=["$\log_a b \cdot \log_b a = 1$","$\log_a b \cdot \log_b c = \log_a c$","$\log_a (b+c)=\log_a b+\log_a c$","$\dfrac{\log_a b}{\log_a c}=\log_c b$"] answer="A,B,D" hint="Use change of base and check which identities are valid." solution="1. True, because $\qquad \log_a b = \dfrac{1}{\log_b a}$

True, because

$\qquad \log_a b \cdot \log_b c = \dfrac{\log b}{\log a}\cdot \dfrac{\log c}{\log b}=\dfrac{\log c}{\log a}=\log_a c$

False, because logarithms do not split over addition.

True, because

$\qquad \dfrac{\log_a b}{\log_a c} = \dfrac{\frac{\log b}{\log a}}{\frac{\log c}{\log a}} = \dfrac{\log b}{\log c} = \log_c b$ Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Simplify $\log_2 3 \cdot \log_3 5 \cdot \log_5 7 \cdot \log_7 16$." answer="4" hint="Use chain cancellation repeatedly." solution="Using the identity $\qquad \log_a b \cdot \log_b c = \log_a c$, we simplify step by step. First, $\qquad \log_2 3 \cdot \log_3 5 = \log_2 5$ Then, $\qquad \log_2 5 \cdot \log_5 7 = \log_2 7$ Then, $\qquad \log_2 7 \cdot \log_7 16 = \log_2 16$ Finally, $\qquad \log_2 16 = 4$ Therefore, the value of the expression is $\boxed{4}$." ::: ---

Summary

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Part 3: Logarithmic equations

Logarithmic Equations

Overview

Logarithmic equations are solved not by memorising isolated tricks, but by controlling three things at every step:

the domain

the logarithm laws

the validity of transformations

In CMI-style algebra, the real challenge is that many formally correct-looking manipulations become invalid if the argument of a logarithm is non-positive or if the base is illegal. So this topic is as much about logical discipline as about computation. ---

Learning Objectives

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Definition and Meaning

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Fundamental Conditions

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Core Formulas

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Most Important Solving Patterns

1. Same base on both sides

If $\qquad \log_a(f(x)) = \log_a(g(x))$ then solve $\qquad f(x)=g(x)$ but only after ensuring $\qquad f(x)>0,\ g(x)>0$ ---

2. Log equal to a number

If $\qquad \log_a(f(x)) = c$ then convert to exponential form: $\qquad f(x) = a^c$ with the restriction $f(x)>0$ automatically necessary. ---

3. Sum or difference of logs

Use product/quotient rules first: $\qquad \log_a M + \log_a N = \log_a(MN)$ $\qquad \log_a M - \log_a N = \log_a\left(\dfrac{M}{N}\right)$ Then reduce to a simpler logarithmic equation. ---

4. Quadratic after substitution

Sometimes a logarithmic expression repeats. Let $\qquad t = \log_a x$ and reduce the equation to a polynomial in $t$. ---

Domain Discipline

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Typical CMI Traps

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Minimal Worked Examples

Example 1 Solve $\qquad \log_2(x-1)=3$ Since the logarithm exists only when $x-1>0$, we need $x>1$. Now convert to exponential form: $\qquad x-1 = 2^3 = 8$ $\qquad x=9$ This satisfies the domain. So the solution is $\boxed{9}$. --- Example 2 Solve $\qquad \log_3(x+1) + \log_3(x-1) = 2$ Domain: $\qquad x+1>0,\ x-1>0 \implies x>1$ Combine the logs: $\qquad \log_3\big((x+1)(x-1)\big)=2$ $\qquad \log_3(x^2-1)=2$ Convert to exponential form: $\qquad x^2-1 = 3^2 = 9$ $\qquad x^2 = 10$ $\qquad x=\pm\sqrt{10}$ From the domain $x>1$, only $x=\sqrt{10}$ is valid. So the solution is $\boxed{\sqrt{10}}$. ---

Strategy for Harder Problems

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Practice Questions

:::question type="MCQ" question="How many real solutions does the equation $\log_2(x-1)=\log_2(5-x)$ have?" options=["0","1","2","3"] answer="B" hint="First write the domain, then compare arguments." solution="For the logarithms to exist, we need $\qquad x-1>0$ and $\qquad 5-x>0$ So $\qquad 1<x<5$ Since the bases are the same and both arguments are positive, $\qquad \log_2(x-1)=\log_2(5-x)$ implies $\qquad x-1=5-x$ So $\qquad 2x=6$ $\qquad x=3$ This lies in the domain $(1,5)$, so it is valid. Hence there is exactly one real solution. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Solve $\log_3(x+1)+\log_3(x-3)=2$. Enter the positive real solution." answer="4" hint="Combine the logarithms into a single logarithm first." solution="The domain is $\qquad x+1>0,\ x-3>0 \implies x>3$ Now combine the logs: $\qquad \log_3\big((x+1)(x-3)\big)=2$ So $\qquad (x+1)(x-3)=3^2=9$ Expand: $\qquad x^2-2x-3=9$ $\qquad x^2-2x-12=0$ Factor: $\qquad (x-4)(x+3)=0$ So the candidates are $\qquad x=4,\ -3$ From the domain $x>3$, only $x=4$ is valid. Therefore the answer is $\boxed{4}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $\log_a u=\log_a v$, then $u=v$ provided $a>0$, $a\ne 1$, $u>0$, $v>0$","$\log_a(u+v)=\log_a u+\log_a v$ for all positive $u,v$","$\log_a 1=0$ for every valid base $a$","If $\log_2 x=3$, then $x=8$"] answer="A,C,D" hint="Recall the valid logarithm laws carefully." solution="1. True. Equality of logs with the same valid base implies equality of positive arguments.

False. There is no logarithm law for sum inside the argument.

True, because $a^0=1$, so $\log_a 1=0$.

True, since $\log_2 x=3$ means $x=2^3=8$.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Solve the equation $\log_2(x^2-5x+6)=1$ over the real numbers." answer="$x=2$ or $x=3$" hint="Convert the logarithmic equation into an algebraic equation and then verify the domain." solution="We first note that the logarithm requires $\qquad x^2-5x+6>0$ Now solve the equation: $\qquad \log_2(x^2-5x+6)=1$ This gives $\qquad x^2-5x+6 = 2^1 = 2$ So $\qquad x^2-5x+4=0$ Factor: $\qquad (x-1)(x-4)=0$ Thus the candidates are $\qquad x=1,\ 4$ Now check in the original expression: for $x=1$, $\qquad x^2-5x+6 = 1-5+6 = 2 > 0$ valid for $x=4$, $\qquad x^2-5x+6 = 16-20+6 = 2 > 0$ valid Hence both solutions are valid. Therefore the solution set is $\boxed{\{1,4\}}$." ::: ---

Summary

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Part 4: Exponential equations

Exponential Equations

Overview

Exponential equations are equations in which the variable appears in the exponent. In CMI-style algebra, these questions are usually not about blind formula use; they test whether you can rewrite bases intelligently, use substitution cleanly, and respect the fact that exponential expressions are always positive for positive bases. ---

Learning Objectives

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Core Idea

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Basic Facts You Must Know

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Method 1: Make the Bases Same

Example 1 Solve $4^{x+1} = 8^{x-1}$. Rewrite both sides in base $2$: $\qquad 4^{x+1} = (2^2)^{x+1} = 2^{2x+2}$ $\qquad 8^{x-1} = (2^3)^{x-1} = 2^{3x-3}$ So, $\qquad 2^{2x+2} = 2^{3x-3}$ Hence, $\qquad 2x+2 = 3x-3$ $\qquad x = 5$ So the solution is $\boxed{5}$. ---

Method 2: Use Substitution

Example 2 Solve $3^x + 3^{-x} = \dfrac{10}{3}$. Let $t = 3^x$. Then $t > 0$ and $3^{-x} = \dfrac{1}{t}$. So, $\qquad t + \dfrac{1}{t} = \dfrac{10}{3}$ Multiply by $3t$: $\qquad 3t^2 + 3 = 10t$ $\qquad 3t^2 - 10t + 3 = 0$ Factor: $\qquad (3t-1)(t-3)=0$ So, $\qquad t = \dfrac{1}{3}$ or $t = 3$ Since $t = 3^x$, $\qquad 3^x = 3 \implies x=1$ $\qquad 3^x = \dfrac{1}{3} \implies x=-1$ Hence the solutions are $\boxed{x=1,-1}$. ---

Method 3: Take Logarithms

Important: Taking logarithms is valid only when both sides are positive. ---

Standard Patterns

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Positivity and Domain Logic

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Solve $9^{x-1}=27^{x-2}$." options=["$x=2$","$x=3$","$x=4$","$x=5$"] answer="D" hint="Rewrite both sides in base $3$." solution="Rewrite each side in base $3$. $\qquad 9^{x-1} = (3^2)^{x-1} = 3^{2x-2}$ $\qquad 27^{x-2} = (3^3)^{x-2} = 3^{3x-6}$ So, $\qquad 3^{2x-2} = 3^{3x-6}$ Hence, $\qquad 2x-2 = 3x-6$ $\qquad x = 4$ Therefore the correct option is $\boxed{D}$." ::: :::question type="NAT" question="If $2^x + 2^{-x} = \dfrac{5}{2}$, find the positive value of $x$." answer="1" hint="Set $t=2^x$." solution="Let $t=2^x$. Then $t>0$ and $2^{-x}=\dfrac{1}{t}$. So, $\qquad t + \dfrac{1}{t} = \dfrac{5}{2}$ Multiply by $2t$: $\qquad 2t^2 + 2 = 5t$ $\qquad 2t^2 - 5t + 2 = 0$ Factor: $\qquad (2t-1)(t-2)=0$ Thus, $\qquad t=\dfrac{1}{2}$ or $t=2$ Since $t=2^x$, we get $\qquad 2^x=\dfrac{1}{2} \implies x=-1$ or $\qquad 2^x=2 \implies x=1$ The positive value of $x$ is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $a>0$ and $a\ne1$, then $a^u=a^v$ implies $u=v$","For every real $x$, $3^x>0$","The equation $2^x=-4$ has a real solution","If $t=5^x$, then $t$ can be negative"] answer="A,B" hint="Use the positivity and one-to-one properties of exponentials." solution="1. True. For $a>0$ and $a\ne1$, the function $a^x$ is one-to-one.

True. For positive base $3$, the expression $3^x$ is always positive.

False. A positive-base exponential can never be negative.

False. If $t=5^x$, then $t>0$ always.

Hence the correct statements are $\boxed{A,B}$." ::: :::question type="SUB" question="Solve the equation $4^x-5\cdot2^x+4=0$." answer="$x=0,2$" hint="Write $4^x$ in terms of $2^x$ and substitute." solution="Write $\qquad 4^x=(2^2)^x=2^{2x}$ Let $\qquad t=2^x$ Then $t>0$ and the equation becomes $\qquad t^2-5t+4=0$ Factor: $\qquad (t-1)(t-4)=0$ So, $\qquad t=1$ or $t=4$ Now return to $x$: $\qquad 2^x=1 \implies x=0$ $\qquad 2^x=4 \implies x=2$ Hence the solutions are $\boxed{x=0,2}$." ::: ---

Summary

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Part 5: Logarithmic inequalities

Logarithmic Inequalities

Overview

Logarithmic inequalities are a favorite algebra topic because they test three things at once: domain, monotonicity, and clean transformation into ordinary inequalities. In most problems, the real challenge is not the algebra itself, but remembering that logarithms are defined only for positive arguments and that the inequality direction depends on the base. ---

Learning Objectives

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Core Definition

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Domain Comes First

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Monotonicity and Inequality Direction

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Standard Logarithm Laws Used in Inequalities

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Method 1: One Logarithm Compared to a Constant

Example 1 Solve $\log_2(x-3) \ge 2$. First, domain: $\qquad x-3>0 \implies x>3$ Since base $2>1$, inequality direction stays the same: $\qquad x-3 \ge 2^2 = 4$ $\qquad x \ge 7$ Combining with the domain gives $\qquad x \ge 7$ So the solution set is $\boxed{[7,\infty)}$. ---

Method 2: Logarithm Compared to Logarithm

Example 2 Solve $\log_3(x+1) \le \log_3(2x-5)$. Domain: $\qquad x+1>0 \implies x>-1$ $\qquad 2x-5>0 \implies x>\dfrac52$ So domain is $\qquad x>\dfrac52$ Since $3>1$, compare arguments directly: $\qquad x+1 \le 2x-5$ $\qquad x \ge 6$ Combining with the domain gives $\qquad x \ge 6$ So the solution set is $\boxed{[6,\infty)}$. ---

Method 3: First Combine the Logarithms

Example 3 Solve $\log_2(x-1) + \log_2(x-3) > 1$. Domain: $\qquad x-1>0,\ x-3>0 \implies x>3$ Combine: $\qquad \log_2\big((x-1)(x-3)\big) > 1$ Since base $2>1$, $\qquad (x-1)(x-3) > 2$ $\qquad x^2-4x+3 > 2$ $\qquad x^2-4x+1 > 0$ The roots are $\qquad x = 2 \pm \sqrt{3}$ So $\qquad x<2-\sqrt3 \quad \text{or} \quad x>2+\sqrt3$ Now apply the domain $x>3$. Therefore the final solution is $\qquad x>2+\sqrt3$ Since $2+\sqrt3<4$, this is equivalent to $\boxed{(3,\infty)}$. ---

Base Less Than 1

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Case Splitting with Quotients

This is often where sign charts become necessary. ---

CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Solve $\log_2(x-1) > 3$." options=["$x>8$","$x>9$","$x\ge 9$","$x\ge 8$"] answer="B" hint="First check the domain, then remove the logarithm." solution="For the logarithm to be defined, we need $\qquad x-1>0 \implies x>1$ Since the base $2>1$, the logarithmic function is increasing, so $\qquad \log_2(x-1) > 3 \iff x-1>2^3=8$ Thus, $\qquad x>9$ This already satisfies the domain. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the least integer satisfying $\log_{1/2}(x-1) \le -2$." answer="5" hint="The base is less than $1$, so the inequality reverses when removing the logarithm." solution="First write the domain: $\qquad x-1>0 \implies x>1$ Since $0<\dfrac12<1$, the logarithmic function is decreasing. Therefore $\qquad \log_{1/2}(x-1) \le -2 \iff x-1 \ge \left(\dfrac12\right)^{-2}$ Now $\qquad \left(\dfrac12\right)^{-2}=4$ So $\qquad x-1 \ge 4 \implies x \ge 5$ Hence the least integer satisfying the inequality is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["For $\log_a(f(x))$ to be defined, we need $f(x)>0$","If $a>1$, then $\log_a x$ is increasing on $(0,\infty)$","If $0<a<1$, then $\log_a u > \log_a v$ implies $u>v$","The inequality $\log_3(x-2)\ge 0$ has solution $x\ge 3$"] answer="A,B,D" hint="Use domain rules and monotonicity." solution="1. True. The argument of a logarithm must be strictly positive.

True. For $a>1$, the logarithm is increasing.

False. If $0<a<1$, the logarithm is decreasing, so the inequality direction reverses. Thus $\log_a u > \log_a v$ implies $u<v$.

True. Since base $3>1$,

$\qquad \log_3(x-2)\ge 0 \iff x-2 \ge 1 \iff x \ge 3$ and this already satisfies the domain $x>2$. Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Solve the inequality $\log_3(x-1)+\log_3(x-4)\ge 2$." answer="$x\ge \dfrac{5+3\sqrt5}{2}$" hint="Write the domain first and then combine the logarithms." solution="First write the domain: $\qquad x-1>0,\ x-4>0 \implies x>4$ Now combine the logarithms: $\qquad \log_3\big((x-1)(x-4)\big)\ge 2$ Since the base $3>1$, compare arguments: $\qquad (x-1)(x-4)\ge 3^2=9$ Expand: $\qquad x^2-5x+4 \ge 9$ $\qquad x^2-5x-5 \ge 0$ The roots of $x^2-5x-5=0$ are $\qquad x=\dfrac{5\pm \sqrt{25+20}}{2}=\dfrac{5\pm 3\sqrt5}{2}$ Since the quadratic opens upward, $\qquad x\le \dfrac{5-3\sqrt5}{2} \quad \text{or} \quad x\ge \dfrac{5+3\sqrt5}{2}$ Now intersect with the domain $x>4$. Hence the only valid part is $\qquad x\ge \dfrac{5+3\sqrt5}{2}$ Therefore the solution set is $\boxed{\left[\dfrac{5+3\sqrt5}{2},\infty\right)}$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="Simplify the expression: $\operatorname{log}_2 (16x^3) - 2\operatorname{log}_2 x^2 + \operatorname{log}_2 (\frac{1}{x})$ for $x>0$." options=["$\operatorname{log}_2 (16/x^2)$","$\operatorname{log}_2 (16/x)$","$\operatorname{log}_2 (16x)$","$\operatorname{log}_2 (16x^2)$"] answer="$\operatorname{log}_2 (16/x^2)$" hint="Apply the logarithm laws for powers, products, and quotients sequentially." solution="
Using the logarithm laws:
$\operatorname{log}_2 (16x^3) - 2\operatorname{log}_2 x^2 + \operatorname{log}_2 (\frac{1}{x})$
$= \operatorname{log}_2 (16x^3) - \operatorname{log}_2 (x^2)^2 + \operatorname{log}_2 (x^{-1})$ (Power rule)
$= \operatorname{log}_2 (16x^3) - \operatorname{log}_2 (x^4) + \operatorname{log}_2 (x^{-1})$
Combine using product and quotient rules:
$= \operatorname{log}_2 \left(\frac{16x^3 \cdot x^{-1}}{x^4}\right)$
$= \operatorname{log}_2 \left(\frac{16x^2}{x^4}\right)$
$= \operatorname{log}_2 \left(\frac{16}{x^2}\right)$
The final answer is $\operatorname{log}_2 (16/x^2)$.
"
:::

:::question type="NAT" question="Find the sum of all possible values of $x$ that satisfy the equation $4^{x+1} - 9 \cdot 2^x + 2 = 0$." answer="-1" hint="Consider a substitution to transform this into a quadratic equation." solution="
Let $y = 2^x$.
Then $4^{x+1} = 4^x \cdot 4^1 = (2^x)^2 \cdot 4 = 4y^2$.
Substitute into the equation:
$4y^2 - 9y + 2 = 0$
This is a quadratic equation. Factor it:
$(4y - 1)(y - 2) = 0$
This yields two possible values for $y$:
$4y - 1 = 0 \implies y = \frac{1}{4}$
$y - 2 = 0 \implies y = 2$
Now substitute back $y = 2^x$:
Case 1: $2^x = \frac{1}{4} \implies 2^x = 2^{-2} \implies x = -2$
Case 2: $2^x = 2 \implies 2^x = 2^1 \implies x = 1$
The possible values for $x$ are $-2$ and $1$.
The sum of all possible values of $x$ is $-2 + 1 = -1$.
"
:::

:::question type="MCQ" question="Determine the solution set for the inequality $\operatorname{log}_{1/2} (x^2 - 3x + 2) \ge -1$." options=["$[0, 1] \cup [2, 3]$","$(0, 1) \cup (2, 3)$","$[0, 1) \cup (2, 3]$","$(-\infty, 1) \cup (2, \infty)$"] answer="$[0, 1) \cup (2, 3]$" hint="First, establish the domain of the logarithm. Then, remember to reverse the inequality sign when dealing with a base between 0 and 1." solution="
First, determine the domain of the logarithm. The argument must be positive:
$x^2 - 3x + 2 > 0$
$(x-1)(x-2) > 0$
This implies $x < 1$ or $x > 2$. This is the domain.

Now, solve the inequality. Since the base of the logarithm is $1/2$ (which is between 0 and 1), we must reverse the inequality sign when converting to exponential form:
$\operatorname{log}_{1/2} (x^2 - 3x + 2) \ge -1$
$x^2 - 3x + 2 \le (1/2)^{-1}$
$x^2 - 3x + 2 \le 2$
$x^2 - 3x \le 0$
$x(x-3) \le 0$
This inequality holds for $0 \le x \le 3$.

Finally, combine the solution with the domain. We need values of $x$ such that ($x < 1$ or $x > 2$) AND ($0 \le x \le 3$).
The intersection is:
$[0, 1)$ (from $0 \le x \le 3$ and $x < 1$)
$\cup$
$(2, 3]$ (from $0 \le x \le 3$ and $x > 2$)
So, the solution set is $[0, 1) \cup (2, 3]$.
"
:::

:::question type="NAT" question="Given $\operatorname{log}_a b = 3$ and $\operatorname{log}_b c = 2$. Find the value of $\operatorname{log}_{a^2} (bc^3)$." answer="10.5" hint="Use the change of base formula and logarithm laws to express everything in terms of $\operatorname{log}_a$." solution="
We are given $\operatorname{log}_a b = 3$ and $\operatorname{log}_b c = 2$.
We need to find $\operatorname{log}_{a^2} (bc^3)$.

Using the change of base formula $\operatorname{log}_X Y = \frac{\operatorname{log}_Z Y}{\operatorname{log}_Z X}$, we can write:
$\operatorname{log}_{a^2} (bc^3) = \frac{\operatorname{log}_a (bc^3)}{\operatorname{log}_a (a^2)}$

Apply logarithm laws to the numerator and denominator:
Numerator: $\operatorname{log}_a (bc^3) = \operatorname{log}_a b + \operatorname{log}_a (c^3) = \operatorname{log}_a b + 3\operatorname{log}_a c$.
Denominator: $\operatorname{log}_a (a^2) = 2 \operatorname{log}_a a = 2 \cdot 1 = 2$.

So, the expression becomes $\frac{\operatorname{log}_a b + 3\operatorname{log}_a c}{2}$.
We know $\operatorname{log}_a b = 3$. We need to find $\operatorname{log}_a c$.
From $\operatorname{log}_b c = 2$, we can use the change of base formula again, expressing it in terms of base $a$:
$\operatorname{log}_b c = \frac{\operatorname{log}_a c}{\operatorname{log}_a b} = 2$.
Substitute $\operatorname{log}_a b = 3$:
$\frac{\operatorname{log}_a c}{3} = 2$
$\operatorname{log}_a c = 2 \cdot 3 = 6$.

Now substitute the values of $\operatorname{log}_a b$ and $\operatorname{log}_a c$ back into the main expression:
$\operatorname{log}_{a^2} (bc^3) = \frac{3 + 3(6)}{2} = \frac{3 + 18}{2} = \frac{21}{2} = 10.5$.
"
:::

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