Powers, Exponents, and Logarithms

Overview

In our study of quantitative aptitude, a firm command of the principles governing powers, exponents, and logarithms is indispensable. These concepts provide the mathematical framework for handling quantities that span vast orders of magnitude, from the infinitesimally small to the astronomically large. While appearing as distinct topics, they are, in fact, two sides of the same coin; logarithms represent the inverse operation of exponentiation, where an expression such as $a^x = b$ is equivalent to $\log_a(b) = x$. A thorough understanding of this intrinsic relationship is crucial for simplifying complex calculations and for developing the numerical intuition required for advanced problem-solving.

This chapter is designed to build that foundational understanding. We will systematically explore the laws of indices, which govern operations such as multiplication, division, and the raising of powers to further powers. Subsequently, we shall introduce the concept of the logarithm, elucidating its fundamental properties and its utility in transforming multiplicative problems into more manageable additive ones. Mastery of these tools is not merely an academic exercise; it is a prerequisite for success in the GATE examination, where questions frequently test the ability to manipulate and solve equations involving these forms efficiently and accurately.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Exponents and Powers | Manipulating expressions using laws of indices. |
| 2 | Logarithms | Properties and applications in solving equations. |

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Learning Objectives

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We now turn our attention to Exponents and Powers...
## Part 1: Exponents and Powers

Introduction

The concepts of exponents and powers are fundamental pillars of numerical computation and algebraic manipulation. An exponent is a mathematical notation that indicates the number of times a quantity, known as the base, is multiplied by itself. A mastery of the principles governing exponents is indispensable for simplifying complex expressions, solving a wide array of equations, and understanding the behavior of functions, particularly those modeling growth and decay phenomena.

In the context of the GATE examination, questions involving exponents are frequently integrated with other algebraic concepts, such as quadratic equations, polynomials, and inequalities. A thorough understanding of the laws of exponents and their application in various problem-solving scenarios is therefore critical. This chapter will systematically present these laws, explore their use in conjunction with algebraic identities, and provide strategies for tackling typical GATE-level problems. We shall focus on the direct application of these principles to solve problems efficiently and accurately.

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Key Concepts

1. Fundamental Laws of Exponents

The manipulation of expressions involving powers is governed by a set of well-defined laws. These rules are the bedrock upon which all further analysis is built and must be committed to memory. Let us consider $a$ and $b$ to be real numbers ($a, b \neq 0$) and $m, n$ to be rational numbers.

Product of Powers: When multiplying two terms with the same base, we add their exponents.

Quotient of Powers: When dividing two terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator.

Power of a Power: When a power is raised to another power, we multiply the exponents.

Power of a Product: The power of a product of two numbers is the product of their individual powers.

Power of a Quotient: The power of a fraction is the power of the numerator divided by the power of the denominator.

Zero Exponent: Any non-zero base raised to the power of zero is equal to 1.

Negative Exponent: A base raised to a negative exponent is equivalent to the reciprocal of the base raised to the positive exponent.

Fractional Exponent (Roots): An exponent of the form $1/n$ denotes the $n$-th root. More generally, $a^{m/n}$ is the $n$-th root of $a^m$.

Worked Example:

Problem: Simplify the expression $\frac{(27)^{2/3} \times (8)^{-1/3}}{(16)^{1/4}}$.

Solution:

Step 1: Express the bases as powers of prime numbers.
We can write $27 = 3^3$, $8 = 2^3$, and $16 = 2^4$.

Step 2: Apply the power of a power rule, $(a^m)^n = a^{mn}$.

Step 3: Simplify the exponents.

Step 4: Apply the quotient of powers rule, $\frac{a^m}{a^n} = a^{m-n}$, and the negative exponent rule, $a^{-n} = 1/a^n$.

Step 5: Compute the final value.

Answer: $\boxed{\frac{9}{4}}$

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2. Algebraic Identities Involving Powers

Exponents frequently appear within larger algebraic expressions. A common problem type in GATE involves the simplification of such expressions using standard algebraic identities. The difference of squares is particularly important.

Worked Example:

Problem: If $(x+1.5)^2 - (x-0.5)^2 = 2x - 1$, find the value of $x$.

Solution:

Step 1: Identify the structure of the left-hand side (LHS) of the equation.
The LHS is in the form $a^2 - b^2$, where $a = (x+1.5)$ and $b = (x-0.5)$.

Step 2: Apply the difference of squares identity, $a^2 - b^2 = (a-b)(a+b)$.

Step 3: Simplify the terms within each bracket.

For the first bracket $(a-b)$:

For the second bracket $(a+b)$:

Step 4: Substitute the simplified terms back into the equation.

Step 5: Solve the resulting linear equation for $x$.

Answer: $\boxed{x = -\frac{3}{2}}$

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3. Polynomials, Roots, and Simplification

A powerful technique for simplifying expressions involves using the definition of a polynomial's root. If $r$ is a root of the equation $P(x)=0$, it means that $P(r)=0$. This relationship can be used to reduce the degree of a more complex expression involving $r$.

Worked Example:

Problem: Let $\alpha$ be a root of the equation $x^2+x-7=0$. Determine the value of the expression $(\alpha-1)(\alpha-2)(\alpha+2)(\alpha+3)$.

Solution:

Step 1: Use the property of the root $\alpha$.
Since $\alpha$ is a root, it satisfies the equation:

This gives us a key substitution:

Step 2: Strategically reorder and group the terms of the expression. We seek to group pairs whose constants sum to the same value. Observe that for $(\alpha-1)$ and $(\alpha+2)$, the sum of constants is $-1+2=1$. For $(\alpha-2)$ and $(\alpha+3)$, the sum is $-2+3=1$. This suggests the following grouping:

Step 3: Expand the grouped pairs.

First pair:

Second pair:

Step 4: Substitute the relationship from Step 1 into these expanded forms.

The first term becomes:

The second term becomes:

Step 5: Calculate the final value of the expression.

The entire expression is the product of these two simplified values:

Answer: $\boxed{5}$

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4. Comparing Growth Rates of Functions

When comparing functions for large values of a variable, the term with the highest power (the highest degree) dominates the function's behavior. A quadratic function will eventually grow faster than a linear one, and a cubic function will eventually outgrow a quadratic one.

t
y

0

T

g(t)

f(t)

Intersection

g(t) > f(t)
f(t) > g(t)

Worked Example:

Problem: Consider two functions $f(x) = 0.5x^2$ and $g(x) = 50x$ for $x > 0$. Find the smallest integer value $T$ such that for all $x > T$, $f(x) > g(x)$.

Solution:

Step 1: Set up the inequality to be solved.
We need to find the range of $x$ for which $f(x) > g(x)$.

Step 2: Solve the inequality.
Since we are given $x > 0$, we can safely divide both sides by $x$ without changing the inequality direction.

Step 3: Isolate $x$.

Step 4: Interpret the result.
The inequality $f(x) > g(x)$ holds for all values of $x$ greater than 100. The question asks for the smallest integer value $T$ such that for all $x > T$, the condition holds. This value is $T = 100$.

Answer: $\boxed{100}$

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="If $(2x - 0.5)^2 - (2x - 2.5)^2 = 3x - 11$, what is the value of $x$?" options=["$-1$","$1$","$-2$","$2$"] answer="-1" hint="Recognize the expression as a difference of squares, $a^2 - b^2 = (a-b)(a+b)$." solution="
Step 1: Apply the difference of squares identity.

Step 2: Simplify the terms in the brackets.

Step 3: Solve the linear equation.

Answer: $\boxed{-1}$
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:::question type="NAT" question="Calculate the value of the expression $\frac{81^{3/4} + 125^{2/3}}{16^{-1/2}}$." answer="208" hint="First, express the bases as powers of prime numbers. Then, use the negative exponent rule to move the denominator to the numerator." solution="
Step 1: Rewrite the bases as powers.
$81 = 3^4$, $125 = 5^3$, $16 = 4^2$.

Step 2: Simplify the exponents using the rule $(a^m)^n = a^{mn}$.

Step 3: Evaluate the powers in the numerator.

Step 4: Apply the negative exponent rule.

Step 5: Compute the final result.

Answer: $\boxed{208}$
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:::question type="MSQ" question="Let $x$ and $y$ be positive real numbers. Which of the following statements are ALWAYS true?" options=["$(x^2 y^3)^4 = x^8 y^{12}$","$\sqrt{x^2 + y^2} = x+y$","$\frac{x^n}{x^{-n}} = x^{2n}$","$x^{-1} + y^{-1} = (x+y)^{-1}$"] answer="A,C" hint="Apply the fundamental laws of exponents to each option. Test the invalid options with simple numerical examples." solution="
Option A: $(x^2 y^3)^4 = x^8 y^{12}$
Using the power of a product rule, $(ab)^n = a^n b^n$:
$(x^2 y^3)^4 = (x^2)^4 (y^3)^4$.
Using the power of a power rule, $(a^m)^n = a^{mn}$:
$(x^2)^4 (y^3)^4 = x^{2 \times 4} y^{3 \times 4} = x^8 y^{12}$.
This statement is correct.

Option B: $\sqrt{x^2 + y^2} = x+y$
This is a common algebraic mistake. Let's test with $x=3, y=4$.
LHS: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
RHS: $3+4=7$.
Since $5 \neq 7$, this statement is incorrect.

Option C: $\frac{x^n}{x^{-n}} = x^{2n}$
Using the quotient of powers rule, $\frac{a^m}{a^n} = a^{m-n}$:
$\frac{x^n}{x^{-n}} = x^{n - (-n)} = x^{n+n} = x^{2n}$.
This statement is correct.

Option D: $x^{-1} + y^{-1} = (x+y)^{-1}$
Using the negative exponent rule, $a^{-1} = 1/a$:
LHS: $x^{-1} + y^{-1} = \frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}$.
RHS: $(x+y)^{-1} = \frac{1}{x+y}$.
Since $\frac{x+y}{xy} \neq \frac{1}{x+y}$ in general, this statement is incorrect.

Therefore, only options A and C are always true.

Answer: $\boxed{A, C}$
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:::question type="MCQ" question="If $r$ is a root of the equation $x^2 - 5x + 2 = 0$, then the value of the expression $r(r-1)(r-4)(r-5)$ is:" options=["$-4$","$-2$","$0$","$2$"] answer="-4" hint="Group the terms strategically to create a common expression involving $r^2 - 5r$." solution="
Step 1: Use the property of the root $r$.
$r^2 - 5r + 2 = 0 \implies r^2 - 5r = -2$.

Step 2: Group the terms.
Group $r$ with $(r-5)$ and $(r-1)$ with $(r-4)$.
$[r(r-5)][(r-1)(r-4)]$.

Step 3: Expand the grouped pairs.
First pair: $r(r-5) = r^2 - 5r$.
Second pair: $(r-1)(r-4) = r^2 - 5r + 4$.

Step 4: Substitute the relationship from Step 1.
The first term is $r^2 - 5r = -2$.
The second term is $(r^2 - 5r) + 4 = -2 + 4 = 2$.

Step 5: Calculate the final value.
The expression is the product: $(-2)(2) = -4$.

Answer: $\boxed{-4}$
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:::

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Summary

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What's Next?

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Part 2: Logarithms

Introduction

The logarithm is a fundamental concept in mathematics that serves as the inverse operation to exponentiation. In essence, if we ask what power a given base must be raised to in order to yield a certain number, the answer is the logarithm. The study of logarithms is indispensable for simplifying calculations that involve multiplication, division, and exponentiation, transforming them into more manageable operations of addition, subtraction, and multiplication, respectively.

For the GATE examination, a firm grasp of logarithmic properties is not merely an exercise in algebraic manipulation. It is a foundational skill required for numerical computation and for understanding more advanced topics in data science, such as information theory, algorithmic complexity (e.g., $O(\log n)$), and the behavior of certain statistical distributions. This chapter will provide a rigorous treatment of the core principles and properties of logarithms, equipping the aspirant with the necessary tools to solve related problems with precision and efficiency. We will focus exclusively on the algebraic properties and problem-solving techniques relevant to the GATE syllabus.

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Key Concepts and Properties of Logarithms

1. The Product Rule

The logarithm of a product of two numbers is equivalent to the sum of their individual logarithms, provided they share the same base. This property allows us to convert a multiplication operation into an addition.

Worked Example:

Problem: Given $\log_{10}(2) \approx 0.301$ and $\log_{10}(3) \approx 0.477$, find the value of $\log_{10}(6)$.

Solution:

Step 1: Express the argument as a product of the given numbers.

We can write $6$ as $2 \times 3$.

Step 2: Apply the Product Rule.

Step 3: Substitute the given values.

Step 4: Compute the final sum.

Answer: The approximate value of $\log_{10}(6)$ is $0.778$.

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2. The Quotient Rule

Analogous to the product rule, the logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator. This rule transforms a division operation into a subtraction.

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3. The Power Rule

The power rule is exceptionally useful for dealing with exponents inside a logarithm. It states that the logarithm of a number raised to a power is the product of the power and the logarithm of the number itself.

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4. Change of Base Formula

It is often necessary to convert a logarithm from one base to another, especially when calculators or tables are only available for a standard base (like 10 or $e$). The change of base formula provides a straightforward method for this conversion.

A particularly useful corollary of this formula is for inverting the base and argument:

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## 5. Fundamental Logarithmic Identities

There are several identities that follow directly from the definition of a logarithm. While simple, they are crucial for simplification and problem-solving.

Logarithm of the Base: The logarithm of the base itself is always 1.

Logarithm of Unity: The logarithm of 1 to any valid base is always 0.

Inverse Property: Raising a base to the power of a logarithm with the same base results in the argument of the logarithm.

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## 6. Solving Logarithmic Equations

A common type of problem in competitive exams involves solving equations containing logarithmic terms. The primary strategy relies on using the properties above to simplify the equation into a specific form.

Worked Example:

Problem: Solve for $x$ in the equation:

Solution:

Step 1: Use the Product Rule to combine the logarithmic terms.
The left side is a sum of two logarithms with the same base. We can combine them into a single logarithm.

Step 2: Simplify the argument.

Step 3: Convert the logarithmic equation to its exponential form.
Using the definition $\log_b(a) = x \iff b^x = a$.

Step 4: Solve the resulting algebraic equation for $x$.

Step 5: Verify the solution(s) in the original equation.
The argument of a logarithm must be positive.

• For $x=3$: $\log_2(3+1) = \log_2(4)$ and $\log_2(3-1) = \log_2(2)$. Both arguments are positive, so $x=3$ is a valid solution.


• For $x=-3$: $\log_2(-3+1) = \log_2(-2)$. The argument is negative. This is undefined. Thus, $x=-3$ is an extraneous solution.

Answer: The only valid solution is $x=3$.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="If $2\log(x) = \log(2) + \log(3x-4)$, what is the value of $x$?" options=["1", "2", "4", "Both 2 and 4"] answer="Both 2 and 4" hint="Use the power rule to handle the coefficient $2$, then apply the product rule on the right side. Finally, equate the arguments." solution="
Step 1: Apply the power rule to the left side of the equation.

Step 2: Apply the product rule to the right side of the equation.

Step 3: Use the principle of logarithmic equality to equate the arguments.

Step 4: Solve the resulting quadratic equation.

This gives two potential solutions: $x=2$ and $x=4$.

Step 5: Verify both solutions in the original equation.
For $x=2$: The arguments are $x=2 > 0$ and $3x-4 = 3(2)-4 = 2 > 0$. Both are valid.
For $x=4$: The arguments are $x=4 > 0$ and $3x-4 = 3(4)-4 = 8 > 0$. Both are valid.

Result: Both $x=2$ and $x=4$ are valid solutions.
Answer: \boxed{\text{Both 2 and 4}}
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:::question type="NAT" question="Calculate the value of $\log_3(27\sqrt{3})$." answer="3.5" hint="Express the argument $27\sqrt{3}$ as a single power of 3." solution="
Step 1: Express the argument in terms of powers of the base 3.

So, $27\sqrt{3} = 3^3 \times 3^{1/2}$.

Step 2: Use the rules of exponents to combine the terms.

Step 3: Substitute this back into the logarithm.

Step 4: Apply the power rule for logarithms.

Step 5: Use the identity $\log_b(b)=1$.

Result: The value is $3.5$.
Answer: \boxed{3.5}
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:::question type="MSQ" question="Which of the following statements are always true for $x, y, b > 0$ and $b \neq 1$?" options=["$\log_b(x) + \log_b(y) = \log_b(x+y)$", "$\log_b(x^y) = y \log_b(x)$", "$(\log_b x)/(\log_b y) = \log_y x$", "$\log_{b^2}(x) = 2 \log_b(x)$"] answer="B,C" hint="Evaluate each option against the fundamental properties of logarithms. Use the change of base formula for options C and D." solution="

• Option A: $\log_b(x) + \log_b(y) = \log_b(x+y)$. This is a common mistake. The correct product rule is $\log_b(x) + \log_b(y) = \log_b(xy)$. So, A is incorrect.

• Option B: $\log_b(x^y) = y \log_b(x)$. This is the definition of the power rule. So, B is correct.

• Option C: $(\log_b x)/(\log_b y) = \log_y x$. Let us use the change of base formula. We can write $\log_y(x)$ with a new base $b$: $\log_y(x) = \frac{\log_b(x)}{\log_b(y)}$. This matches the expression given. So, C is correct.

• Option D: $\log_{b^2}(x) = 2 \log_b(x)$. Let's use the change of base formula. Let $L = \log_{b^2}(x)$. The exponential form is $(b^2)^L = x$, which is $b^{2L} = x$. The logarithmic form is $\log_b(x) = 2L$. Therefore, $L = \frac{1}{2}\log_b(x)$. The statement claims the result is $2 \log_b(x)$. So, D is incorrect.

Result: The correct statements are B and C.
Answer: \boxed{\text{B, C}}
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:::question type="MCQ" question="If $\ln(x^2 - y^2) - \ln(x-y) = \ln(8)$, what is the value of $x+y$?" options=["2", "4", "8", "16"] answer="8" hint="Use the quotient rule on the left side to combine the terms. Simplify the resulting argument." solution="
Step 1: Apply the quotient rule to the left side of the equation.

Step 2: Simplify the argument inside the logarithm. The numerator is a difference of squares.

Step 3: Substitute the simplified argument back into the equation.

Step 4: Apply the principle of logarithmic equality.

Result: The value of $x+y$ is 8.
Answer: \boxed{8}
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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="If $2^x = 3^y = 12^z$, which of the following relations is correct?" options=["$\frac{1}{z} = \frac{2}{x} + \frac{1}{y}$","$\frac{1}{x} = \frac{1}{y} + \frac{2}{z}$","$\frac{1}{y} = \frac{1}{z} + \frac{1}{x}$","$\frac{1}{z} = \frac{1}{x} - \frac{2}{y}$"] answer="A" hint="Let $2^x = 3^y = 12^z = k$. Express 2, 3, and 12 in terms of $k$ and their respective exponents. Then, use the relationship $12 = 4 \times 3 = 2^2 \times 3$." solution="
Let us assume $2^x = 3^y = 12^z = k$ for some constant $k$.

From this assumption, we can express the bases in terms of $k$:

• $2^x = k \implies 2 = k^{1/x}$


• $3^y = k \implies 3 = k^{1/y}$


• $12^z = k \implies 12 = k^{1/z}$

We know the fundamental relationship between the bases: $12 = 4 \times 3 = 2^2 \times 3$.

Step 1: Substitute the expressions in terms of $k$ into this relationship.

Step 2: Apply the laws of exponents.

Step 3: Since the bases are equal, equate the exponents.

Thus, the correct relation is given by option A.
Answer: \boxed{\text{A}}
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:::question type="NAT" question="Find the value of the expression $81^{(\frac{1}{\log_5 3})} + 27^{\log_9 36} + 3^{(\frac{4}{\log_7 9})}$." answer="890" hint="Use the change of base property $\frac{1}{\log_b a} = \log_a b$ for the first and third terms. For the second term, express 27 and 9 as powers of 3 and simplify." solution="
We shall evaluate each term of the expression separately.

Term 1: $81^{(\frac{1}{\log_5 3})}$
Using the identity $\frac{1}{\log_b a} = \log_a b$, we have:

Expressing 81 as a power of 3:

Using the identity $a^{\log_a x} = x$:

Term 2: $27^{\log_9 36}$
Let's simplify the exponent first. Using the change of base formula:

So the term becomes:

Term 3: $3^{(\frac{4}{\log_7 9})}$
Again, using $\frac{1}{\log_b a} = \log_a b$:

Expressing 9 as a power of 3 in the logarithm's base:

So the term becomes:

Final Calculation:
The value of the expression is the sum of the three terms:

Answer: \boxed{890}
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:::question type="NAT" question="If $\log_x (x-1) + \log_x (x+1) = \log_x (x+5)$, and $x > 1$, what is the value of $x$?" answer="3" hint="Use the product rule for logarithms to combine the terms on the left-hand side. Then, equate the arguments of the logarithms on both sides to form a solvable equation." solution="
The given equation is $\log_x (x-1) + \log_x (x+1) = \log_x (x+5)$.
The domain for the logarithms requires $x > 0$, $x \neq 1$, $x-1 > 0$, $x+1 > 0$, and $x+5 > 0$. The strictest of these conditions is $x > 1$, which is given in the problem statement.

Step 1: Use the product rule for logarithms to combine the terms on the left side.

This simplifies to:

Step 2: Since the logarithms on both sides have the same base, equate their arguments.

Step 3: Rearrange the terms to form a quadratic equation and solve.

This gives two possible solutions for $x$: $x=3$ and $x=-2$.

Step 4: Verify the solutions against the domain constraint.
The domain constraint is $x > 1$.
The solution $x=-2$ is extraneous as it falls outside this domain.
The solution $x=3$ is valid because $3 > 1$.

Answer: \boxed{3}
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What's Next?