Algebra Practice Questions — JAMB UTME — Page 5

Question 81Question →

If $\log_4 x + \log_2 y = \frac{5}{2}$ and $3^x \cdot 9^{-y} = 81$ , what is the value of $x + y$ ?

6

8

10

12

Show answer & explanation

Answer: 10

Answer

The value of

x + y

is

10

.

Converting

\log_4 x

to base

2

gives

\frac{1}{2}\log_2 x

, leadings to

x y^2 = 32

. Simplifying

3^x \cdot 9^{-y} = 81

gives

x - 2y = 4

. Solving these simultaneously gives

x = 8

and

y = 2

, whose sum is

10

.

Step-by-Step Solution

1

Simplify the logarithmic equation using the change of base formula.

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x

. Substituting this gives

\frac{1}{2}\log_2 x + \log_2 y = \frac{5}{2} \implies \log_2 x + 2\log_2 y = 5

.

Converting logarithms to a common base of

2

allows combining terms.

2

Combine logarithmic terms and express as an algebraic relation.

\log_2(x y^2) = 5 \implies x y^2 = 2^5 = 32

.

Applying the logarithmic laws

k\log_b a = \log_b(a^k)

and

\log_b M + \log_b N = \log_b(MN)

.

3

Simplify the exponential equation.

3^x \cdot (3^2)^{-y} = 3^4 \implies 3^{x - 2y} = 3^4 \implies x - 2y = 4

.

Expressing both sides with base

3

allows equating exponents.

4

Solve the system of equations for

x

and

y

.

From

x = 4 + 2y

, substitute into

x y^2 = 32

:

(4 + 2y)y^2 = 32 \implies 2y^3 + 4y^2 - 32 = 0 \implies y^3 + 2y^2 - 16 = 0

. Testing positive factors yields

y = 2

. Then

x = 4 + 2(2) = 8

.

Simultaneous substitution yields the unique real positive solutions

x = 8

and

y = 2

.

5

Calculate

x + y

.

x + y = 8 + 2 = 10

.

Adding the computed values of

x

and

y

gives the required sum.

Key Concept

Simultaneous Exponential and Logarithmic Systems

Estimated Time:2m 0s

Question 82Question →

How many integer values of $x$ satisfy both the linear inequality $\frac{3 - 2x}{5} \ge \frac{x + 4}{2}$ and the quadratic inequality $x^2 + 4x - 5 \le 0$ ?

4

3

5

2

Show answer & explanation

Answer: 4

Answer

4

Solving the linear inequality yields

x \le -\frac{14}{9} \approx -1.56

. Solving the quadratic inequality gives

-5 \le x \le 1

. The overlap between both sets is

-5 \le x \le -\frac{14}{9}

. The integers falling within this interval are

-5, -4, -3

, and

-2

, making 4 valid integer solutions in total.

Step-by-Step Solution

1

Solve the linear inequality

\frac{3 - 2x}{5} \ge \frac{x + 4}{2}

.

2(3 - 2x) \ge 5(x + 4) \implies 6 - 4x \ge 5x + 20 \implies -9x \ge 14 \implies x \le -\frac{14}{9} \approx -1.56

.

Clear denominators by multiplying by 10 and reverse the inequality sign when dividing both sides by

-9

.

2

Solve the quadratic inequality

x^2 + 4x - 5 \le 0

.

(x + 5)(x - 1) \le 0 \implies -5 \le x \le 1

.

Factorize the quadratic expression to find critical points at

x = -5

and

x = 1

. The region where the product is non-positive is between the roots.

3

Determine the intersection of the two solution sets.

-5 \le x \le -\frac{14}{9}

.

Combine the conditions

x \le -1.56

and

-5 \le x \le 1

to find the set of values satisfying both inequalities simultaneously.

4

Count the integer values within the intersection set

[-5, -1.56]

.

The integers are

-5, -4, -3, -2

, giving a total of 4 integers.

Identify all whole numbers within the combined solution interval.

Key Concept

Simultaneous Linear and Quadratic Inequalities

Question 83Question →

When the polynomial $P(x) = x^3 - ax^2 + bx - 6$ is divided by $(x - 1)$ , the remainder is $-4$ . If $(x - 2)$ is a factor of $P(x)$ , what is the value of $a + b$ ?

5

1

-1

-3

Show answer & explanation

Answer: $5$

Answer

The value of

a + b

is

5

.

Using the Remainder Theorem with

P(1) = -4

gives the equation

-a + b = 1

. Using the Factor Theorem with

P(2) = 0

gives

2a - b = 1

. Solving this simultaneous system gives

a = 2

and

b = 3

, leading to

a + b = 5

.

Step-by-Step Solution

1

Apply the Remainder Theorem for division by

(x - 1)

.

-a + b = 1

According to the Remainder Theorem,

P(1) = -4

. Substituting

x = 1

into

P(x)

yields

1^3 - a(1)^2 + b(1) - 6 = -4 \implies -a + b = 1

.

2

Apply the Factor Theorem for the factor

(x - 2)

.

2a - b = 1

Since

(x - 2)

is a factor,

P(2) = 0

. Substituting

x = 2

into

P(x)

yields

2^3 - a(2)^2 + b(2) - 6 = 0 \implies 8 - 4a + 2b - 6 = 0 \implies 2a - b = 1

.

3

Solve the system of linear equations simultaneously.

a = 2, b = 3

Adding the two equations

(-a + b) + (2a - b) = 1 + 1

yields

a = 2

. Substituting

a = 2

back into

-a + b = 1

gives

b = 3

.

4

Calculate the required value

a + b

.

5

a + b = 2 + 3 = 5

.

Key Concept

Polynomial Factor and Remainder Theorems

Estimated Time:1m 30s

Question 84Question →

When the polynomial $P(x) = 2x^4 + ax^3 + bx^2 - 5x + 6$ is divided by $(x - 2)(x + 1)$ , the remainder is $6x + 8$ . What is the value of $a - b$ ?

11

-11

1

-1

Show answer & explanation

Answer: 11

Answer

11

According to the Remainder Theorem, dividing

P(x)

by

(x - 2)(x + 1)

leaves a remainder

R(x) = 6x + 8

. Substituting

x = -1

gives

R(-1) = 2

and

P(-1) = 2 - a + b + 5 + 6 = 13 - a + b

. Equating

13 - a + b = 2

gives

-a + b = -11

, which directly leads to

a - b = 11

.

Step-by-Step Solution

1

Apply the Remainder Theorem for linear factors of the divisor

(x - 2)(x + 1)

.

Since the divisor is

(x - 2)(x + 1)

, the roots of the divisor are

x = 2

and

x = -1

. The remainder function is

R(x) = 6x + 8

, so

P(2) = R(2)

and

P(-1) = R(-1)

.

By the Polynomial Division Algorithm,

P(x) = (x - 2)(x + 1)Q(x) + R(x)

.

2

Evaluate

R(x)

and

P(x)

at

x = -1

.

R(-1) = 6(-1) + 8 = 2

.

P(-1) = 2(-1)^4 + a(-1)^3 + b(-1)^2 - 5(-1) + 6 = 2 - a + b + 5 + 6 = 13 - a + b

.
Setting

P(-1) = R(-1)

gives

13 - a + b = 2

.

Substituting

x = -1

eliminates the quotient term since

(-1 + 1) = 0

.

3

Rearrange the equation to solve for

a - b

.

13 - a + b = 2 \implies -a + b = -11 \implies a - b = 11

.

Multiplying both sides of

-a + b = -11

by

-1

gives

a - b = 11

.

Key Concept

Remainder Theorem for Composite Linear Divisors

Question 85Question →

Given that the matrix $A = \begin{pmatrix} x & 2 \\ 5 & x - 3 \end{pmatrix}$ is a singular matrix, what is the positive value of $x$ ?

2

3

5

10

Show answer & explanation

Answer: $5$

Answer

The positive value of

x

is

5

.

For a matrix to be singular, its determinant must be zero. Calculating the determinant of

A

yields

\det(A) = x(x - 3) - (2)(5) = x^2 - 3x - 10

. Setting this to zero and factoring gives

(x - 5)(x + 2) = 0

, leading to

x = 5

or

x = -2

. The positive solution is

5

.

Step-by-Step Solution

1

Set the determinant of the matrix equal to zero.

\det(A) = (x)(x - 3) - (2)(5) = 0

By definition, a matrix is singular if and only if its determinant is zero.

2

Expand and simplify the algebraic equation.

x^2 - 3x - 10 = 0

Expanding

x(x - 3)

gives

x^2 - 3x

, and subtracting

10

forms a standard quadratic equation.

3

Factor the quadratic equation to solve for

x

.

(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2

The factors of

-10

that sum to

-3

are

-5

and

+2

.

4

Select the positive value requested by the question.

x = 5

The question specifically asks for the positive value of

x

.

Key Concept

Singular Matrices and Determinants

Question 86Question →

The second term of a geometric progression (G.P.) is $6$ and its fifth term is $48$ . What is the sum of the first $6$ terms of the progression?

189

93

96

378

Show answer & explanation

Answer: 189

Answer

189

Using the geometric progression terms

T_2 = a r = 6

and

T_5 = a r^4 = 48

, dividing

T_5

by

T_2

gives

r^3 = 8

, so

r = 2

. Substituting

r = 2

into

a r = 6

yields

a = 3

. The sum of the first 6 terms is calculated as

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 189

.

Step-by-Step Solution

1

Set up equations using the

n^{\text{th}}

term formula for a G.P.,

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

Relate given terms to the first term

a

and common ratio

r

.

2

Divide the equation for

T_5

by the equation for

T_2

to find the common ratio

r

.

\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2

Eliminate the variable

a

to solve for

r

.

3

Substitute

r = 2

back into

a r = 6

to find the first term

a

.

a(2) = 6 \implies a = 3

Determine the first term of the progression.

4

Calculate the sum of the first

6

terms using

S_n = \frac{a(r^n - 1)}{r - 1}

.

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(64 - 1) = 3(63) = 189

Apply the sum formula for a finite geometric progression.

Key Concept

Geometric Progression (G.P.)

n^{\text{th}}

term and sum of finite terms

Question 87Question →

If the matrix $M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}$ is a singular matrix, what is the value of $x$ ?

1

-1

2

0

Show answer & explanation

Answer: $1$

Answer

The value of

x

is

1

.

A matrix is singular when its determinant equals zero. Evaluating the determinant of

\begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}

gives

(2x)(6) - (4)(3) = 12x - 12

. Setting

12x - 12 = 0

leads directly to

x = 1

.

Step-by-Step Solution

1

Recall the condition for a matrix to be singular.

A matrix is singular if and only if its determinant is equal to zero, so

\det(M) = 0

.

Singular matrices have no inverse because their determinant is zero.

2

Calculate the determinant of matrix

M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}

.

\det(M) = (2x)(6) - (4)(3) = 12x - 12.

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is

ad - bc

.

3

Set the determinant equal to zero and solve for

x

.

12x - 12 = 0 \implies 12x = 12 \implies x = 1

.

Solving the linear equation yields the required parameter value.

Key Concept

Condition for Singular Matrix and 2x2 Determinant Evaluation

Estimated Time:45s

Question 88Question →

For the matrix $M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}$ , the determinant of $M$ is equal to $4$ . What is the positive value of $k$ ?

Show answer & explanation

Answer: 2

Answer

The positive value of

k

is

2

.

Expanding the determinant along the second row yields

\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2

. Setting this equal to

4

yields

2k^2 - k - 6 = 0

, which factors into

(2k+3)(k-2)=0

. The positive value is

2

.

Step-by-Step Solution

1

Evaluate the determinant of

M

in terms of

k

using row 2 cofactor expansion

\det(M) = 2k^2 - k - 2

Expanding along the second row gives

-2(6 - 5) + k(2k - 1) - 0 = -2 + 2k^2 - k

.

2

Set the determinant equal to the given value 4 and rearrange into standard quadratic form

2k^2 - k - 6 = 0

Subtracting 4 from both sides gives

2k^2 - k - 6 = 0

.

3

Solve the quadratic equation by factorization

k = -1.5 \text{ or } k = 2

Factoring

(2k + 3)(k - 2) = 0

gives roots

k = -1.5

and

k = 2

.

4

Select the positive root

k = 2

The question specifically asks for the positive value of

k

.

Key Concept

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.

Question 89Question →

Given the matrices $A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 2 & y \end{pmatrix}$ , if $AB = BA$ , what is the value of $x + y$ ?

-3

10

-1

1

Show answer & explanation

Answer: $-3$

Answer

The value of

x + y

is

-3

.

Evaluating the matrix products

AB

and

BA

using standard matrix multiplication rules yields

AB = \begin{pmatrix} 2+2x & 8+xy \\ 5 & -4+3y \end{pmatrix}

and

BA = \begin{pmatrix} -2 & x+12 \\ 4-y & 2x+3y \end{pmatrix}

. Equating entry-by-entry gives

2+2x = -2 \implies x = -2

and

5 = 4-y \implies y = -1

. Adding these together gives

x + y = -3

.

Step-by-Step Solution

1

Compute the product matrix

AB

AB = \begin{pmatrix} 2(1) + x(2) & 2(4) + x(y) \\ -1(1) + 3(2) & -1(4) + 3(y) \end{pmatrix} = \begin{pmatrix} 2 + 2x & 8 + xy \\ 5 & -4 + 3y \end{pmatrix}

Matrix multiplication requires taking the dot product of rows from the first matrix and columns from the second matrix.

2

Compute the product matrix

BA

BA = \begin{pmatrix} 1(2) + 4(-1) & 1(x) + 4(3) \\ 2(2) + y(-1) & 2(x) + y(3) \end{pmatrix} = \begin{pmatrix} -2 & x + 12 \\ 4 - y & 2x + 3y \end{pmatrix}

Evaluate

BA

by multiplying rows of

B

by columns of

A

.

3

Equate corresponding entries of

AB

and

BA

since

AB = BA

From row 1, col 1:

2 + 2x = -2 \implies 2x = -4 \implies x = -2

.
From row 2, col 1:

5 = 4 - y \implies y = -1

.

Two matrices are equal if and only if all corresponding entries are equal.

4

Verify consistency on remaining entries and calculate

x + y

Row 1, col 2 check:

8 + (-2)(-1) = 10

and

-2 + 12 = 10

.
Row 2, col 2 check:

-4 + 3(-1) = -7

and

2(-2) + 3(-1) = -7

.
Sum:

x + y = -2 + (-1) = -3

.

Verifying consistency ensures the system of equations has a valid unique solution.

Key Concept

Matrix Multiplication Commutativity and Matrix Equality

Estimated Time:2m 30s

Question 90Question →

What is the positive value of $y$ that satisfies the simultaneous equations $x - y = 2$ and $x^2 - y^2 = 12$ ?

2

4

6

8

Show answer & explanation

Answer: 2

Answer

The positive value of y is 2.

Factoring

x^2 - y^2

as

(x - y)(x + y)

gives

2(x + y) = 12

, leading to

x + y = 6

. Subtracting

x - y = 2

from

x + y = 6

isolates

2y = 4

, giving

y = 2

.

Step-by-Step Solution

1

Factorize the quadratic difference of squares equation.

x^2 - y^2 = (x - y)(x + y) = 12

Difference of two squares identity allows linear substitution.

2

Substitute

x - y = 2

into the factorized expression.

2(x + y) = 12 \implies x + y = 6

Simplifies the quadratic system into a second linear equation.

3

Solve the system of linear equations

x + y = 6

and

x - y = 2

for

y

.

(x + y) - (x - y) = 6 - 2 \implies 2y = 4 \implies y = 2

Subtracting the two linear equations eliminates

x

and directly isolates

y

.

Key Concept

Simultaneous Linear and Quadratic Equations via Difference of Squares Substitution

Estimated Time:45s

Question 91Question →

Find the product of all real solutions to the exponential equation $9^{x+1} - 28 \cdot 3^x + 3 = 0$ .

Show answer & explanation

Answer: -2

Answer

The product of all real solutions is -2.

Using the index laws

9^{x+1} = 9 \cdot (3^x)^2

, let

u = 3^x

. The equation reduces to the quadratic

9u^2 - 28u + 3 = 0

, which factors as

(9u - 1)(u - 3) = 0

. This gives

u = 1/9

or

u = 3

. Solving

3^x = 1/9

yields

x = -2

, and solving

3^x = 3

yields

x = 1

. The product of these solutions is

(-2) \times 1 = -2

.

Step-by-Step Solution

1

Express

9^{x+1}

in terms of

3^x

9^{x+1} = 9^1 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2

Apply index laws

a^{m+n} = a^m \cdot a^n

and

(a^m)^n = (a^n)^m

to establish a common base of 3.

2

Substitute

u = 3^x

to form a quadratic equation

9u^2 - 28u + 3 = 0

Transform the exponential equation into a standard quadratic algebraic equation.

3

Solve the quadratic equation for

u

(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9}

or

u = 3

Factorize the quadratic expression to determine its roots.

4

Substitute back

u = 3^x

to solve for

x

3^x = 3^{-2} \implies x = -2

, and

3^x = 3^1 \implies x = 1

Equate exponents with matching bases to find all valid real solutions for

x

.

5

Find the product of the two solutions

(-2) \times 1 = -2

Calculate the required mathematical product of the solutions.

Key Concept

Solving exponential equations reducible to quadratic form using laws of indices

Question 92Question →

If $\log_4 x + \log_x 16 = 3$ , what is the sum of the possible values of $x$ ?

20

12

48

3

Show answer & explanation

Answer: 20

Answer

The sum of the possible values of

x

is 20.

Applying the change of base rule converts

\log_x 16

to

\frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

. Setting

y = \log_4 x

transforms the equation into

y + \frac{2}{y} = 3

, which rearranges to

y^2 - 3y + 2 = 0

. The roots of this quadratic equation are

y = 1

and

y = 2

. Converting back to

x

yields

x = 4^1 = 4

and

x = 4^2 = 16

. Summing these solutions gives

4 + 16 = 20

.

Step-by-Step Solution

1

Apply the change of base formula to express

\log_x 16

in base 4.

\log_x 16 = \frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

.

Logarithmic bases must be unified to combine terms.

2

Substitute

y = \log_4 x

into the original equation.

y + \frac{2}{y} = 3

.

Simplifies the equation into a quadratic form in terms of

y

.

3

Multiply by

y

and solve the quadratic equation

y^2 - 3y + 2 = 0

.

(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2

.

Finds the exponential power values.

4

Solve for

x

using

x = 4^y

and calculate the sum.

For

y = 1

,

x = 4^1 = 4

; for

y = 2

,

x = 4^2 = 16

. Sum =

4 + 16 = 20

.

Converts back from logarithmic space to solve for

x

and finds the requested sum.

Key Concept

Change of base rule and solving logarithmic quadratic equations

Question 93Question →

If $8^{x - 1} = 32$ , what is the value of $x$ ?

2

\frac{8}{3}

5

\frac{2}{3}

Show answer & explanation

Answer: $\frac{8}{3}$

Answer

The value of

x

is

\frac{8}{3}

.

To solve

8^{x - 1} = 32

, rewrite both sides using a base of

2

:

(2^3)^{x - 1} = 2^5

. Applying the index law

(a^m)^n = a^{mn}

yields

2^{3x - 3} = 2^5

. Since the bases are identical, set the exponents equal to each other:

3x - 3 = 5

. Solving for

x

gives

3x = 8

, so

x = \frac{8}{3}

.

Step-by-Step Solution

1

Express both numbers as powers of a common base,

2

.

8 = 2^3

and

32 = 2^5

, so

(2^3)^{x - 1} = 2^5

.

Exponential equations with different bases are easiest to solve by expressing terms with the same base.

2

Apply the law of indices

(a^m)^n = a^{mn}

to simplify the left-hand side.

2^{3(x - 1)} = 2^5

, which expands to

2^{3x - 3} = 2^5

.

Multiplying the inner exponent by the outer exponent removes parentheses.

3

Equate the exponents since the bases are equal.

3x - 3 = 5

.

If

a^P = a^Q

for

a > 0

and

a \neq 1

, then

P = Q

.

4

Solve the linear equation for

x

.

3x = 5 + 3 \implies 3x = 8 \implies x = \frac{8}{3}

.

Isolate

x

by adding

3

to both sides and dividing by

3

.

Key Concept

Solving Exponential Equations using Base Conversion

Estimated Time:1m 0s

Question 94Question →

What is the numerical value of the expression $\log_2(\log_3 81) + \log_5\left(\frac{1}{125}\right) + 4^{\log_2 3}$ ?

Show answer & explanation

Answer: 8

Answer

8

Evaluating each term individually: the nested logarithm

\log_2(\log_3 81) = \log_2 4 = 2

; the reciprocal log argument

\log_5(1/125) = -3

; and the exponential log power

4^{\log_2 3} = (2^{\log_2 3})^2 = 3^2 = 9

. Combining these gives

2 - 3 + 9 = 8

.

Step-by-Step Solution

1

Evaluate the first component

\log_2(\log_3 81)

2

Since

81 = 3^4

, the inner expression

\log_3 81 = 4

, and subsequently

\log_2 4 = 2

.

2

Evaluate the second component

\log_5\left(\frac{1}{125}\right)

-3

Since

\frac{1}{125} = 5^{-3}

, applying the power law of logarithms gives

-3

.

3

Evaluate the third component

4^{\log_2 3}

9

Rewrite

4

as

2^2

to obtain

(2^{\log_2 3})^2 = 3^2 = 9

using the fundamental identity

a^{\log_a b} = b

.

4

Sum the results of the three components

8

Calculate

2 + (-3) + 9 = 8

.

Key Concept

Properties of logarithms including change of power, negative exponents, and logarithm exponentiation identities

Question 95Question →

The sum to infinity of a geometric progression (G.P.) with positive terms is $18$ , and the sum of its first two terms is $16$ . What is the first term of the progression?

Show answer & explanation

Answer: 12

Answer

The first term of the progression is 12.

Using the sum to infinity formula

S_{\infty} = \frac{a}{1 - r} = 18

, we express the first term as

a = 18(1 - r)

. Combining this with the sum of the first two terms

S_2 = a(1 + r) = 16

yields

18(1 - r)(1 + r) = 16 \implies 18(1 - r^2) = 16

. Solving for

r

gives

r^2 = \frac{1}{9}

, so

r = \frac{1}{3}

for a sequence with positive terms. Substituting

r = \frac{1}{3}

back into

a = 18(1 - r)

gives

a = 12

.

Step-by-Step Solution

1

Express the sum to infinity in terms of the first term

a

and common ratio

r

.

a = 18(1 - r)

The sum to infinity formula for a convergent G.P. is

S_{\infty} = \frac{a}{1 - r}

.

2

Write the expression for the sum of the first two terms.

a(1 + r) = 16

The sum of the first two terms is

T_1 + T_2 = a + ar = a(1 + r)

.

3

Substitute

a = 18(1 - r)

into the sum of the first two terms equation.

18(1 - r^2) = 16

Applying the difference of two squares identity

(1 - r)(1 + r) = 1 - r^2

.

4

Solve for the common ratio

r

.

r = \frac{1}{3}

Rearranging gives

1 - r^2 = \frac{8}{9} \implies r^2 = \frac{1}{9}

. Since all terms are positive,

r

must be positive.

5

Calculate the first term

a

.

a = 12

Substitute

r = \frac{1}{3}

into

a = 18(1 - r)

to get

a = 18 \times \frac{2}{3} = 12

.

Key Concept

Geometric Progression sum to infinity and partial sums

Estimated Time:1m 30s

Question 96Question →

What is the solution set for the linear inequality $\frac{1 - 4x}{3} \le \frac{x + 7}{2}$ ?

x \ge -\frac{19}{11}

x \le -\frac{19}{11}

x \ge \frac{19}{11}

x \le \frac{23}{5}

Show answer & explanation

Answer: $x \ge -\frac{19}{11}$

Answer

x \ge -\frac{19}{11}

Multiplying the inequality by 6 clears denominators to give

2(1 - 4x) \le 3(x + 7)

. Expanding gives

2 - 8x \le 3x + 21

, which simplifies to

-11x \le 19

. Dividing both sides by

-11

flips the inequality sign, yielding

x \ge -\frac{19}{11}

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the inequality by the lowest common multiple of 3 and 2, which is 6.

6 \cdot \left(\frac{1 - 4x}{3}\right) \le 6 \cdot \left(\frac{x + 7}{2}\right) \implies 2(1 - 4x) \le 3(x + 7)

Eliminating fractions simplifies the algebraic manipulation.

2

Expand both sides by distributing the factors.

2 - 8x \le 3x + 21

Removing parentheses allows grouping of like terms.

3

Rearrange terms by collecting terms containing

x

on the left and constants on the right.

-8x - 3x \le 21 - 2 \implies -11x \le 19

Isolating the variable term prepares for the final division step.

4

Divide both sides by

-11

and reverse the direction of the inequality sign.

x \ge -\frac{19}{11}

Dividing or multiplying an inequality by a negative real number requires flipping the inequality sign.

Key Concept

Solving linear inequalities involving fractions and reversing the inequality sign when dividing by a negative number.

Estimated Time:1m 30s

Question 97Question →

If $\log_3 x + \log_9 x + \log_{27} x = \frac{11}{2}$ , what is the value of $x$ ?

27

9

3

81

Show answer & explanation

Answer: $27$

Answer

The value of

x

is

27

.

Applying the change of base identity yields

\log_9 x = \frac{1}{2}\log_3 x

and

\log_{27} x = \frac{1}{3}\log_3 x

. Combining like terms gives

\frac{11}{6}\log_3 x = \frac{11}{2}

, which reduces to

\log_3 x = 3

. Expressing in exponential form yields

x = 3^3 = 27

.

Step-by-Step Solution

1

Express all logarithms in base 3 using the change of base rule

\log_{a^k} x = \frac{1}{k} \log_a x

.

\log_9 x = \frac{1}{2} \log_3 x \quad \text{and} \quad \log_{27} x = \frac{1}{3} \log_3 x

Converting all terms to a common base allows them to be combined algebraically.

2

Substitute these expressions back into the original equation.

\log_3 x + \frac{1}{2} \log_3 x + \frac{1}{3} \log_3 x = \frac{11}{2}

This creates a single linear equation in terms of

\log_3 x

.

3

Combine the coefficients of

\log_3 x

.

\left(1 + \frac{1}{2} + \frac{1}{3}\right) \log_3 x = \frac{6 + 3 + 2}{6} \log_3 x = \frac{11}{6} \log_3 x = \frac{11}{2}

Adding the fractions gives a single coefficient.

4

Solve for

\log_3 x

and evaluate

x

.

\log_3 x = \frac{11}{2} \times \frac{6}{11} = 3 \implies x = 3^3 = 27

Converting from logarithmic to exponential form gives the value of

x

.

Key Concept

Change of base formula for logarithms:

\log_{b^k} x = \frac{1}{k} \log_b x

Estimated Time:1m 30s

Question 98Question →

Find the positive integer value of $x$ that satisfies the logarithmic equation $x^{\log_3 x} = 81x^3$ .

Show answer & explanation

Answer: 81

Answer

The positive integer value of

x

is 81.

Taking the logarithm base 3 on both sides transforms

x^{\log_3 x} = 81x^3

into

(\log_3 x)^2 = 4 + 3\log_3 x

. Substituting

u = \log_3 x

yields the quadratic equation

u^2 - 3u - 4 = 0

, which factors as

(u-4)(u+1) = 0

. The roots are

u = 4

(

x = 3^4 = 81

) and

u = -1

(

x = 3^{-1} = 1/3

). Since

x

must be a positive integer, the correct value is 81.

Step-by-Step Solution

1

Take logarithm base 3 of both sides of the equation

x^{\log_3 x} = 81x^3

\log_3(x^{\log_3 x}) = \log_3(81x^3)

Taking the logarithm with base 3 allows us to simplify the exponent containing

\log_3 x

.

2

Apply logarithmic identities

\log_b(a^k) = k \log_b a

and

\log_b(mn) = \log_b m + \log_b n

(\log_3 x)^2 = \log_3 81 + 3\log_3 x = 4 + 3\log_3 x

Expanding the products and powers reduces the equation into a single logarithmic variable

\log_3 x

.

3

Substitute

u = \log_3 x

to create a quadratic equation

u^2 - 3u - 4 = 0

Setting

u = \log_3 x

converts the equation into standard quadratic form.

4

Factor the quadratic expression

(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1

Factoring allows us to find all possible real values for

u

.

5

Convert back to

x

using

x = 3^u

and select the positive integer root

x = 3^4 = 81

or

x = 3^{-1} = \frac{1}{3}

. The positive integer solution is

x = 81

.

The question specifically requests the positive integer solution, eliminating

x = \frac{1}{3}

.

Key Concept

Solving equations with variable exponents by taking logarithms and reducing to a quadratic form.

Question 99Question →

When the polynomial $P(x) = 2x^3 + 3x^2 - px + q$ is divided by $(x - 1)$ , the remainder is $3$ . Given that $(x + 2)$ is a factor of $P(x)$ , what is the value of $p + q$ ?

2

4

6

-2

Show answer & explanation

Answer: 2

Answer

The value of

p + q

is

2

.

According to the Remainder Theorem,

P(1) = 3

gives

2(1)^3 + 3(1)^2 - p(1) + q = 3

, which simplifies to

p - q = 2

. According to the Factor Theorem,

(x + 2)

being a factor means

P(-2) = 0

, giving

2(-2)^3 + 3(-2)^2 - p(-2) + q = 0

, which simplifies to

2p + q = 4

. Solving these equations together gives

p = 2

and

q = 0

. Summing them yields

p + q = 2

.

Step-by-Step Solution

1

Apply the Remainder Theorem for divisor

(x - 1)

P(1) = 2(1)^3 + 3(1)^2 - p(1) + q = 3 \implies 5 - p + q = 3 \implies p - q = 2

By the Remainder Theorem, dividing

P(x)

by

(x - a)

leaves a remainder equal to

P(a)

.

2

Apply the Factor Theorem for factor

(x + 2)

P(-2) = 2(-2)^3 + 3(-2)^2 - p(-2) + q = 0 \implies -16 + 12 + 2p + q = 0 \implies 2p + q = 4

By the Factor Theorem, if

(x - a)

is a factor of

P(x)

, then

P(a) = 0

. Here

a = -2

.

3

Solve the simultaneous linear equations for

p

and

q

Adding

p - q = 2

and

2p + q = 4

yields

3p = 6 \implies p = 2

. Substituting

p = 2

into

p - q = 2

gives

q = 0

.

Eliminating

q

allows direct calculation of

p

, followed by back-substitution for

q

.

4

Calculate the target expression

p + q

p + q = 2 + 0 = 2

Combine the values of

p

and

q

to obtain the required sum.

Key Concept

Factor and Remainder Theorems for Polynomials

Question 100Question →

If the matrix $A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & k & 1 \\ 3 & 1 & 4 \end{pmatrix}$ is singular, what is the value of $k$ ?

\frac{3}{2}

-\frac{3}{2}

\frac{7}{2}

-\frac{1}{2}

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

\frac{3}{2}

A matrix is singular if its determinant equals zero. Expanding the determinant of matrix

A

along its first row gives

1(4k - 1) - 0 + 2(2 - 3k) = -2k + 3

. Setting

-2k + 3 = 0

yields

k = \frac{3}{2}

.

Step-by-Step Solution

1

Apply the condition for a singular matrix

A matrix is singular when its determinant equals zero, so

\det(A) = 0

.

By definition, square matrices with zero determinant are singular.

2

Expand

\det(A)

along the first row

\det(A) = 1(4k - 1) - 0(8 - 3) + 2(2 - 3k) = 4k - 1 + 4 - 6k = -2k + 3

.

Laplace expansion along the first row simplifies computation due to the zero entry.

3

Solve for

k

-2k + 3 = 0 \implies 2k = 3 \implies k = \frac{3}{2}

.

Isolating the variable gives the required value of

k

.

Key Concept

Singular Matrix Condition and 3x3 Determinant Expansion

Algebra

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