Algebra

239 soru

Soru 81Soru

If log4x+log2y=52\log_4 x + \log_2 y = \frac{5}{2} and 3x9y=813^x \cdot 9^{-y} = 81, what is the value of x+yx + y?

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Cevap: 10

Cevap

The value of x+yx + y is 1010.
Converting log4x\log_4 x to base 22 gives 12log2x\frac{1}{2}\log_2 x, leadings to xy2=32x y^2 = 32. Simplifying 3x9y=813^x \cdot 9^{-y} = 81 gives x2y=4x - 2y = 4. Solving these simultaneously gives x=8x = 8 and y=2y = 2, whose sum is 1010.

Adım Adım Çözüm

1
Simplify the logarithmic equation using the change of base formula.
log4x=log2xlog24=12log2x\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x. Substituting this gives 12log2x+log2y=52    log2x+2log2y=5\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 22 allows combining terms.
2
Combine logarithmic terms and express as an algebraic relation.
log2(xy2)=5    xy2=25=32\log_2(x y^2) = 5 \implies x y^2 = 2^5 = 32.
Applying the logarithmic laws klogba=logb(ak)k\log_b a = \log_b(a^k) and logbM+logbN=logb(MN)\log_b M + \log_b N = \log_b(MN).
3
Simplify the exponential equation.
3x(32)y=34    3x2y=34    x2y=43^x \cdot (3^2)^{-y} = 3^4 \implies 3^{x - 2y} = 3^4 \implies x - 2y = 4.
Expressing both sides with base 33 allows equating exponents.
4
Solve the system of equations for xx and yy.
From x=4+2yx = 4 + 2y, substitute into xy2=32x y^2 = 32: (4+2y)y2=32    2y3+4y232=0    y3+2y216=0(4 + 2y)y^2 = 32 \implies 2y^3 + 4y^2 - 32 = 0 \implies y^3 + 2y^2 - 16 = 0. Testing positive factors yields y=2y = 2. Then x=4+2(2)=8x = 4 + 2(2) = 8.
Simultaneous substitution yields the unique real positive solutions x=8x = 8 and y=2y = 2.
5
Calculate x+yx + y.
x+y=8+2=10x + y = 8 + 2 = 10.
Adding the computed values of xx and yy gives the required sum.

Anahtar Kavram

Simultaneous Exponential and Logarithmic Systems
Tahmini Süre:2m 0s
Soru 82Soru

How many integer values of xx satisfy both the linear inequality 32x5x+42\frac{3 - 2x}{5} \ge \frac{x + 4}{2} and the quadratic inequality x2+4x50x^2 + 4x - 5 \le 0?

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Cevap: 4

Cevap

4
Solving the linear inequality yields x1491.56x \le -\frac{14}{9} \approx -1.56. Solving the quadratic inequality gives 5x1-5 \le x \le 1. The overlap between both sets is 5x149-5 \le x \le -\frac{14}{9}. The integers falling within this interval are 5,4,3-5, -4, -3, and 2-2, making 4 valid integer solutions in total.

Adım Adım Çözüm

1
Solve the linear inequality 32x5x+42\frac{3 - 2x}{5} \ge \frac{x + 4}{2}.
2(32x)5(x+4)    64x5x+20    9x14    x1491.562(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 9-9.
2
Solve the quadratic inequality x2+4x50x^2 + 4x - 5 \le 0.
(x+5)(x1)0    5x1(x + 5)(x - 1) \le 0 \implies -5 \le x \le 1.
Factorize the quadratic expression to find critical points at x=5x = -5 and x=1x = 1. The region where the product is non-positive is between the roots.
3
Determine the intersection of the two solution sets.
5x149-5 \le x \le -\frac{14}{9}.
Combine the conditions x1.56x \le -1.56 and 5x1-5 \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][-5, -1.56].
The integers are 5,4,3,2-5, -4, -3, -2, giving a total of 4 integers.
Identify all whole numbers within the combined solution interval.

Anahtar Kavram

Simultaneous Linear and Quadratic Inequalities
Soru 83Soru

When the polynomial P(x)=x3ax2+bx6P(x) = x^3 - ax^2 + bx - 6 is divided by (x1)(x - 1), the remainder is 4-4. If (x2)(x - 2) is a factor of P(x)P(x), what is the value of a+ba + b?

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Cevap: 55

Cevap

The value of a+ba + b is 55.
Using the Remainder Theorem with P(1)=4P(1) = -4 gives the equation a+b=1-a + b = 1. Using the Factor Theorem with P(2)=0P(2) = 0 gives 2ab=12a - b = 1. Solving this simultaneous system gives a=2a = 2 and b=3b = 3, leading to a+b=5a + b = 5.

Adım Adım Çözüm

1
Apply the Remainder Theorem for division by (x1)(x - 1).
a+b=1-a + b = 1
According to the Remainder Theorem, P(1)=4P(1) = -4. Substituting x=1x = 1 into P(x)P(x) yields 13a(1)2+b(1)6=4    a+b=11^3 - a(1)^2 + b(1) - 6 = -4 \implies -a + b = 1.
2
Apply the Factor Theorem for the factor (x2)(x - 2).
2ab=12a - b = 1
Since (x2)(x - 2) is a factor, P(2)=0P(2) = 0. Substituting x=2x = 2 into P(x)P(x) yields 23a(2)2+b(2)6=0    84a+2b6=0    2ab=12^3 - 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=3a = 2, b = 3
Adding the two equations (a+b)+(2ab)=1+1(-a + b) + (2a - b) = 1 + 1 yields a=2a = 2. Substituting a=2a = 2 back into a+b=1-a + b = 1 gives b=3b = 3.
4
Calculate the required value a+ba + b.
55
a+b=2+3=5a + b = 2 + 3 = 5.

Anahtar Kavram

Polynomial Factor and Remainder Theorems
Tahmini Süre:1m 30s
Soru 84Soru

When the polynomial P(x)=2x4+ax3+bx25x+6P(x) = 2x^4 + ax^3 + bx^2 - 5x + 6 is divided by (x2)(x+1)(x - 2)(x + 1), the remainder is 6x+86x + 8. What is the value of aba - b?

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Cevap: 11

Cevap

11
According to the Remainder Theorem, dividing P(x)P(x) by (x2)(x+1)(x - 2)(x + 1) leaves a remainder R(x)=6x+8R(x) = 6x + 8. Substituting x=1x = -1 gives R(1)=2R(-1) = 2 and P(1)=2a+b+5+6=13a+bP(-1) = 2 - a + b + 5 + 6 = 13 - a + b. Equating 13a+b=213 - a + b = 2 gives a+b=11-a + b = -11, which directly leads to ab=11a - b = 11.

Adım Adım Çözüm

1
Apply the Remainder Theorem for linear factors of the divisor (x2)(x+1)(x - 2)(x + 1).
Since the divisor is (x2)(x+1)(x - 2)(x + 1), the roots of the divisor are x=2x = 2 and x=1x = -1. The remainder function is R(x)=6x+8R(x) = 6x + 8, so P(2)=R(2)P(2) = R(2) and P(1)=R(1)P(-1) = R(-1).
By the Polynomial Division Algorithm, P(x)=(x2)(x+1)Q(x)+R(x)P(x) = (x - 2)(x + 1)Q(x) + R(x).
2
Evaluate R(x)R(x) and P(x)P(x) at x=1x = -1.
R(1)=6(1)+8=2R(-1) = 6(-1) + 8 = 2.
P(1)=2(1)4+a(1)3+b(1)25(1)+6=2a+b+5+6=13a+bP(-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)P(-1) = R(-1) gives 13a+b=213 - a + b = 2.
Substituting x=1x = -1 eliminates the quotient term since (1+1)=0(-1 + 1) = 0.
3
Rearrange the equation to solve for aba - b.
13a+b=2    a+b=11    ab=1113 - a + b = 2 \implies -a + b = -11 \implies a - b = 11.
Multiplying both sides of a+b=11-a + b = -11 by 1-1 gives ab=11a - b = 11.

Anahtar Kavram

Remainder Theorem for Composite Linear Divisors
Soru 85Soru

Given that the matrix A=(x25x3)A = \begin{pmatrix} x & 2 \\ 5 & x - 3 \end{pmatrix} is a singular matrix, what is the positive value of xx?

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Cevap: 55

Cevap

The positive value of xx is 55.
For a matrix to be singular, its determinant must be zero. Calculating the determinant of AA yields det(A)=x(x3)(2)(5)=x23x10\det(A) = x(x - 3) - (2)(5) = x^2 - 3x - 10. Setting this to zero and factoring gives (x5)(x+2)=0(x - 5)(x + 2) = 0, leading to x=5x = 5 or x=2x = -2. The positive solution is 55.

Adım Adım Çözüm

1
Set the determinant of the matrix equal to zero.
det(A)=(x)(x3)(2)(5)=0\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.
x23x10=0x^2 - 3x - 10 = 0
Expanding x(x3)x(x - 3) gives x23xx^2 - 3x, and subtracting 1010 forms a standard quadratic equation.
3
Factor the quadratic equation to solve for xx.
(x5)(x+2)=0    x=5 or x=2(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2
The factors of 10-10 that sum to 3-3 are 5-5 and +2+2.
4
Select the positive value requested by the question.
x=5x = 5
The question specifically asks for the positive value of xx.

Anahtar Kavram

Singular Matrices and Determinants
Soru 86Soru

The second term of a geometric progression (G.P.) is 66 and its fifth term is 4848. What is the sum of the first 66 terms of the progression?

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Cevap: 189

Cevap

189
Using the geometric progression terms T2=ar=6T_2 = a r = 6 and T5=ar4=48T_5 = a r^4 = 48, dividing T5T_5 by T2T_2 gives r3=8r^3 = 8, so r=2r = 2. Substituting r=2r = 2 into ar=6a r = 6 yields a=3a = 3. The sum of the first 6 terms is calculated as S6=3(261)21=189S_6 = \frac{3(2^6 - 1)}{2 - 1} = 189.

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1
Set up equations using the nthn^{\text{th}} term formula for a G.P., Tn=arn1T_n = a r^{n-1}.
T2=ar=6T_2 = a r = 6 and T5=ar4=48T_5 = a r^4 = 48
Relate given terms to the first term aa and common ratio rr.
2
Divide the equation for T5T_5 by the equation for T2T_2 to find the common ratio rr.
\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2
Eliminate the variable aa to solve for rr.
3
Substitute r=2r = 2 back into ar=6a r = 6 to find the first term aa.
a(2) = 6 \implies a = 3
Determine the first term of the progression.
4
Calculate the sum of the first 66 terms using Sn=a(rn1)r1S_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.

Anahtar Kavram

Geometric Progression (G.P.) nthn^{\text{th}} term and sum of finite terms
Soru 87Soru

If the matrix M=(2x436)M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix} is a singular matrix, what is the value of xx?

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Cevap: 11

Cevap

The value of xx is 11.
A matrix is singular when its determinant equals zero. Evaluating the determinant of (2x436)\begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix} gives (2x)(6)(4)(3)=12x12(2x)(6) - (4)(3) = 12x - 12. Setting 12x12=012x - 12 = 0 leads directly to x=1x = 1.

Adım Adım Çözüm

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\det(M) = 0.
Singular matrices have no inverse because their determinant is zero.
2
Calculate the determinant of matrix M=(2x436)M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}.
\det(M) = (2x)(6) - (4)(3) = 12x - 12.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is adbcad - bc.
3
Set the determinant equal to zero and solve for xx.
12x12=0    12x=12    x=112x - 12 = 0 \implies 12x = 12 \implies x = 1.
Solving the linear equation yields the required parameter value.

Anahtar Kavram

Condition for Singular Matrix and 2x2 Determinant Evaluation
Tahmini Süre:45s
Soru 88Soru

For the matrix M=(k312k0152)M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}, the determinant of MM is equal to 44. What is the positive value of kk?

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Cevap: 2

Cevap

The positive value of kk is 22.
Expanding the determinant along the second row yields det(M)=2(65)+k(2k1)=2k2k2\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2. Setting this equal to 44 yields 2k2k6=02k^2 - k - 6 = 0, which factors into (2k+3)(k2)=0(2k+3)(k-2)=0. The positive value is 22.

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1
Evaluate the determinant of MM in terms of kk using row 2 cofactor expansion
\det(M) = 2k^2 - k - 2
Expanding along the second row gives 2(65)+k(2k1)0=2+2k2k-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 2k2k6=02k^2 - k - 6 = 0.
3
Solve the quadratic equation by factorization
k = -1.5 \text{ or } k = 2
Factoring (2k+3)(k2)=0(2k + 3)(k - 2) = 0 gives roots k=1.5k = -1.5 and k=2k = 2.
4
Select the positive root
k = 2
The question specifically asks for the positive value of kk.

Anahtar Kavram

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.
Soru 89Soru

Given the matrices A=(2x13)A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} and B=(142y)B = \begin{pmatrix} 1 & 4 \\ 2 & y \end{pmatrix}, if AB=BAAB = BA, what is the value of x+yx + y?

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Cevap: 3-3

Cevap

The value of x+yx + y is 3-3.
Evaluating the matrix products ABAB and BABA using standard matrix multiplication rules yields AB=(2+2x8+xy54+3y)AB = \begin{pmatrix} 2+2x & 8+xy \\ 5 & -4+3y \end{pmatrix} and BA=(2x+124y2x+3y)BA = \begin{pmatrix} -2 & x+12 \\ 4-y & 2x+3y \end{pmatrix}. Equating entry-by-entry gives 2+2x=2    x=22+2x = -2 \implies x = -2 and 5=4y    y=15 = 4-y \implies y = -1. Adding these together gives x+y=3x + y = -3.

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1
Compute the product matrix ABAB
AB=(2(1)+x(2)2(4)+x(y)1(1)+3(2)1(4)+3(y))=(2+2x8+xy54+3y)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 BABA
BA=(1(2)+4(1)1(x)+4(3)2(2)+y(1)2(x)+y(3))=(2x+124y2x+3y)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 BABA by multiplying rows of BB by columns of AA.
3
Equate corresponding entries of ABAB and BABA since AB=BAAB = BA
From row 1, col 1: 2+2x=2    2x=4    x=22 + 2x = -2 \implies 2x = -4 \implies x = -2.
From row 2, col 1: 5=4y    y=15 = 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+yx + y
Row 1, col 2 check: 8+(2)(1)=108 + (-2)(-1) = 10 and 2+12=10-2 + 12 = 10.
Row 2, col 2 check: 4+3(1)=7-4 + 3(-1) = -7 and 2(2)+3(1)=72(-2) + 3(-1) = -7.
Sum: x+y=2+(1)=3x + y = -2 + (-1) = -3.
Verifying consistency ensures the system of equations has a valid unique solution.

Anahtar Kavram

Matrix Multiplication Commutativity and Matrix Equality
Tahmini Süre:2m 30s
Soru 90Soru

What is the positive value of yy that satisfies the simultaneous equations xy=2x - y = 2 and x2y2=12x^2 - y^2 = 12?

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Cevap: 2

Cevap

The positive value of y is 2.
Factoring x2y2x^2 - y^2 as (xy)(x+y)(x - y)(x + y) gives 2(x+y)=122(x + y) = 12, leading to x+y=6x + y = 6. Subtracting xy=2x - y = 2 from x+y=6x + y = 6 isolates 2y=42y = 4, giving y=2y = 2.

Adım Adım Çözüm

1
Factorize the quadratic difference of squares equation.
x2y2=(xy)(x+y)=12x^2 - y^2 = (x - y)(x + y) = 12
Difference of two squares identity allows linear substitution.
2
Substitute xy=2x - y = 2 into the factorized expression.
2(x+y)=12    x+y=62(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=6x + y = 6 and xy=2x - y = 2 for yy.
(x+y)(xy)=62    2y=4    y=2(x + y) - (x - y) = 6 - 2 \implies 2y = 4 \implies y = 2
Subtracting the two linear equations eliminates xx and directly isolates yy.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations via Difference of Squares Substitution
Tahmini Süre:45s
Soru 91Soru

Find the product of all real solutions to the exponential equation 9x+1283x+3=09^{x+1} - 28 \cdot 3^x + 3 = 0.

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Cevap: -2

Cevap

The product of all real solutions is -2.
Using the index laws 9x+1=9(3x)29^{x+1} = 9 \cdot (3^x)^2, let u=3xu = 3^x. The equation reduces to the quadratic 9u228u+3=09u^2 - 28u + 3 = 0, which factors as (9u1)(u3)=0(9u - 1)(u - 3) = 0. This gives u=1/9u = 1/9 or u=3u = 3. Solving 3x=1/93^x = 1/9 yields x=2x = -2, and solving 3x=33^x = 3 yields x=1x = 1. The product of these solutions is (2)×1=2(-2) \times 1 = -2.

Adım Adım Çözüm

1
Express 9x+19^{x+1} in terms of 3x3^x
9x+1=919x=9(32)x=9(3x)29^{x+1} = 9^1 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2
Apply index laws am+n=amana^{m+n} = a^m \cdot a^n and (am)n=(an)m(a^m)^n = (a^n)^m to establish a common base of 3.
2
Substitute u=3xu = 3^x to form a quadratic equation
9u228u+3=09u^2 - 28u + 3 = 0
Transform the exponential equation into a standard quadratic algebraic equation.
3
Solve the quadratic equation for uu
(9u1)(u3)=0    u=19(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9} or u=3u = 3
Factorize the quadratic expression to determine its roots.
4
Substitute back u=3xu = 3^x to solve for xx
3x=32    x=23^x = 3^{-2} \implies x = -2, and 3x=31    x=13^x = 3^1 \implies x = 1
Equate exponents with matching bases to find all valid real solutions for xx.
5
Find the product of the two solutions
(2)×1=2(-2) \times 1 = -2
Calculate the required mathematical product of the solutions.

Anahtar Kavram

Solving exponential equations reducible to quadratic form using laws of indices
Soru 92Soru

If log4x+logx16=3\log_4 x + \log_x 16 = 3, what is the sum of the possible values of xx?

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Cevap: 20

Cevap

The sum of the possible values of xx is 20.
Applying the change of base rule converts logx16\log_x 16 to log416log4x=2log4x\frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}. Setting y=log4xy = \log_4 x transforms the equation into y+2y=3y + \frac{2}{y} = 3, which rearranges to y23y+2=0y^2 - 3y + 2 = 0. The roots of this quadratic equation are y=1y = 1 and y=2y = 2. Converting back to xx yields x=41=4x = 4^1 = 4 and x=42=16x = 4^2 = 16. Summing these solutions gives 4+16=204 + 16 = 20.

Adım Adım Çözüm

1
Apply the change of base formula to express logx16\log_x 16 in base 4.
logx16=log416log4x=2log4x\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=log4xy = \log_4 x into the original equation.
y+2y=3y + \frac{2}{y} = 3.
Simplifies the equation into a quadratic form in terms of yy.
3
Multiply by yy and solve the quadratic equation y23y+2=0y^2 - 3y + 2 = 0.
(y1)(y2)=0    y=1 or y=2(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2.
Finds the exponential power values.
4
Solve for xx using x=4yx = 4^y and calculate the sum.
For y=1y = 1, x=41=4x = 4^1 = 4; for y=2y = 2, x=42=16x = 4^2 = 16. Sum = 4+16=204 + 16 = 20.
Converts back from logarithmic space to solve for xx and finds the requested sum.

Anahtar Kavram

Change of base rule and solving logarithmic quadratic equations
Soru 93Soru

If 8x1=328^{x - 1} = 32, what is the value of xx?

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Cevap: 83\frac{8}{3}

Cevap

The value of xx is 83\frac{8}{3}.
To solve 8x1=328^{x - 1} = 32, rewrite both sides using a base of 22: (23)x1=25(2^3)^{x - 1} = 2^5. Applying the index law (am)n=amn(a^m)^n = a^{mn} yields 23x3=252^{3x - 3} = 2^5. Since the bases are identical, set the exponents equal to each other: 3x3=53x - 3 = 5. Solving for xx gives 3x=83x = 8, so x=83x = \frac{8}{3}.

Adım Adım Çözüm

1
Express both numbers as powers of a common base, 22.
8=238 = 2^3 and 32=2532 = 2^5, so (23)x1=25(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 (am)n=amn(a^m)^n = a^{mn} to simplify the left-hand side.
23(x1)=252^{3(x - 1)} = 2^5, which expands to 23x3=252^{3x - 3} = 2^5.
Multiplying the inner exponent by the outer exponent removes parentheses.
3
Equate the exponents since the bases are equal.
3x3=53x - 3 = 5.
If aP=aQa^P = a^Q for a>0a > 0 and a1a \neq 1, then P=QP = Q.
4
Solve the linear equation for xx.
3x=5+3    3x=8    x=833x = 5 + 3 \implies 3x = 8 \implies x = \frac{8}{3}.
Isolate xx by adding 33 to both sides and dividing by 33.

Anahtar Kavram

Solving Exponential Equations using Base Conversion
Tahmini Süre:1m 0s
Soru 94Soru

What is the numerical value of the expression log2(log381)+log5(1125)+4log23\log_2(\log_3 81) + \log_5\left(\frac{1}{125}\right) + 4^{\log_2 3}?

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Cevap: 8

Cevap

8
Evaluating each term individually: the nested logarithm log2(log381)=log24=2\log_2(\log_3 81) = \log_2 4 = 2; the reciprocal log argument log5(1/125)=3\log_5(1/125) = -3; and the exponential log power 4log23=(2log23)2=32=94^{\log_2 3} = (2^{\log_2 3})^2 = 3^2 = 9. Combining these gives 23+9=82 - 3 + 9 = 8.

Adım Adım Çözüm

1
Evaluate the first component log2(log381)\log_2(\log_3 81)
2
Since 81=3481 = 3^4, the inner expression log381=4\log_3 81 = 4, and subsequently log24=2\log_2 4 = 2.
2
Evaluate the second component log5(1125)\log_5\left(\frac{1}{125}\right)
-3
Since 1125=53\frac{1}{125} = 5^{-3}, applying the power law of logarithms gives 3-3.
3
Evaluate the third component 4log234^{\log_2 3}
9
Rewrite 44 as 222^2 to obtain (2log23)2=32=9(2^{\log_2 3})^2 = 3^2 = 9 using the fundamental identity alogab=ba^{\log_a b} = b.
4
Sum the results of the three components
8
Calculate 2+(3)+9=82 + (-3) + 9 = 8.

Anahtar Kavram

Properties of logarithms including change of power, negative exponents, and logarithm exponentiation identities
Soru 95Soru

The sum to infinity of a geometric progression (G.P.) with positive terms is 1818, and the sum of its first two terms is 1616. What is the first term of the progression?

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Cevap: 12

Cevap

The first term of the progression is 12.
Using the sum to infinity formula S=a1r=18S_{\infty} = \frac{a}{1 - r} = 18, we express the first term as a=18(1r)a = 18(1 - r). Combining this with the sum of the first two terms S2=a(1+r)=16S_2 = a(1 + r) = 16 yields 18(1r)(1+r)=16    18(1r2)=1618(1 - r)(1 + r) = 16 \implies 18(1 - r^2) = 16. Solving for rr gives r2=19r^2 = \frac{1}{9}, so r=13r = \frac{1}{3} for a sequence with positive terms. Substituting r=13r = \frac{1}{3} back into a=18(1r)a = 18(1 - r) gives a=12a = 12.

Adım Adım Çözüm

1
Express the sum to infinity in terms of the first term aa and common ratio rr.
a=18(1r)a = 18(1 - r)
The sum to infinity formula for a convergent G.P. is S=a1rS_{\infty} = \frac{a}{1 - r}.
2
Write the expression for the sum of the first two terms.
a(1+r)=16a(1 + r) = 16
The sum of the first two terms is T1+T2=a+ar=a(1+r)T_1 + T_2 = a + ar = a(1 + r).
3
Substitute a=18(1r)a = 18(1 - r) into the sum of the first two terms equation.
18(1r2)=1618(1 - r^2) = 16
Applying the difference of two squares identity (1r)(1+r)=1r2(1 - r)(1 + r) = 1 - r^2.
4
Solve for the common ratio rr.
r=13r = \frac{1}{3}
Rearranging gives 1r2=89    r2=191 - r^2 = \frac{8}{9} \implies r^2 = \frac{1}{9}. Since all terms are positive, rr must be positive.
5
Calculate the first term aa.
a=12a = 12
Substitute r=13r = \frac{1}{3} into a=18(1r)a = 18(1 - r) to get a=18×23=12a = 18 \times \frac{2}{3} = 12.

Anahtar Kavram

Geometric Progression sum to infinity and partial sums
Tahmini Süre:1m 30s
Soru 96Soru

What is the solution set for the linear inequality 14x3x+72\frac{1 - 4x}{3} \le \frac{x + 7}{2}?

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Cevap: x1911x \ge -\frac{19}{11}

Cevap

x1911x \ge -\frac{19}{11}
Multiplying the inequality by 6 clears denominators to give 2(14x)3(x+7)2(1 - 4x) \le 3(x + 7). Expanding gives 28x3x+212 - 8x \le 3x + 21, which simplifies to 11x19-11x \le 19. Dividing both sides by 11-11 flips the inequality sign, yielding x1911x \ge -\frac{19}{11}.

Adım Adım Çözüm

1
Clear the denominators by multiplying both sides of the inequality by the lowest common multiple of 3 and 2, which is 6.
6(14x3)6(x+72)    2(14x)3(x+7)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.
28x3x+212 - 8x \le 3x + 21
Removing parentheses allows grouping of like terms.
3
Rearrange terms by collecting terms containing xx on the left and constants on the right.
8x3x212    11x19-8x - 3x \le 21 - 2 \implies -11x \le 19
Isolating the variable term prepares for the final division step.
4
Divide both sides by 11-11 and reverse the direction of the inequality sign.
x1911x \ge -\frac{19}{11}
Dividing or multiplying an inequality by a negative real number requires flipping the inequality sign.

Anahtar Kavram

Solving linear inequalities involving fractions and reversing the inequality sign when dividing by a negative number.
Tahmini Süre:1m 30s
Soru 97Soru

If log3x+log9x+log27x=112\log_3 x + \log_9 x + \log_{27} x = \frac{11}{2}, what is the value of xx?

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Cevap: 2727

Cevap

The value of xx is 2727.
Applying the change of base identity yields log9x=12log3x\log_9 x = \frac{1}{2}\log_3 x and log27x=13log3x\log_{27} x = \frac{1}{3}\log_3 x. Combining like terms gives 116log3x=112\frac{11}{6}\log_3 x = \frac{11}{2}, which reduces to log3x=3\log_3 x = 3. Expressing in exponential form yields x=33=27x = 3^3 = 27.

Adım Adım Çözüm

1
Express all logarithms in base 3 using the change of base rule logakx=1klogax\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 log3x\log_3 x.
3
Combine the coefficients of log3x\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 log3x\log_3 x and evaluate xx.
\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 xx.

Anahtar Kavram

Change of base formula for logarithms: logbkx=1klogbx\log_{b^k} x = \frac{1}{k} \log_b x
Tahmini Süre:1m 30s
Soru 98Soru

Find the positive integer value of xx that satisfies the logarithmic equation xlog3x=81x3x^{\log_3 x} = 81x^3.

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Cevap: 81

Cevap

The positive integer value of xx is 81.
Taking the logarithm base 3 on both sides transforms xlog3x=81x3x^{\log_3 x} = 81x^3 into (log3x)2=4+3log3x(\log_3 x)^2 = 4 + 3\log_3 x. Substituting u=log3xu = \log_3 x yields the quadratic equation u23u4=0u^2 - 3u - 4 = 0, which factors as (u4)(u+1)=0(u-4)(u+1) = 0. The roots are u=4u = 4 (x=34=81x = 3^4 = 81) and u=1u = -1 (x=31=1/3x = 3^{-1} = 1/3). Since xx must be a positive integer, the correct value is 81.

Adım Adım Çözüm

1
Take logarithm base 3 of both sides of the equation xlog3x=81x3x^{\log_3 x} = 81x^3
log3(xlog3x)=log3(81x3)\log_3(x^{\log_3 x}) = \log_3(81x^3)
Taking the logarithm with base 3 allows us to simplify the exponent containing log3x\log_3 x.
2
Apply logarithmic identities logb(ak)=klogba\log_b(a^k) = k \log_b a and logb(mn)=logbm+logbn\log_b(mn) = \log_b m + \log_b n
(log3x)2=log381+3log3x=4+3log3x(\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 log3x\log_3 x.
3
Substitute u=log3xu = \log_3 x to create a quadratic equation
u23u4=0u^2 - 3u - 4 = 0
Setting u=log3xu = \log_3 x converts the equation into standard quadratic form.
4
Factor the quadratic expression
(u4)(u+1)=0    u=4 or u=1(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1
Factoring allows us to find all possible real values for uu.
5
Convert back to xx using x=3ux = 3^u and select the positive integer root
x=34=81x = 3^4 = 81 or x=31=13x = 3^{-1} = \frac{1}{3}. The positive integer solution is x=81x = 81.
The question specifically requests the positive integer solution, eliminating x=13x = \frac{1}{3}.

Anahtar Kavram

Solving equations with variable exponents by taking logarithms and reducing to a quadratic form.
Soru 99Soru

When the polynomial P(x)=2x3+3x2px+qP(x) = 2x^3 + 3x^2 - px + q is divided by (x1)(x - 1), the remainder is 33. Given that (x+2)(x + 2) is a factor of P(x)P(x), what is the value of p+qp + q?

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Cevap: 2

Cevap

The value of p+qp + q is 22.
According to the Remainder Theorem, P(1)=3P(1) = 3 gives 2(1)3+3(1)2p(1)+q=32(1)^3 + 3(1)^2 - p(1) + q = 3, which simplifies to pq=2p - q = 2. According to the Factor Theorem, (x+2)(x + 2) being a factor means P(2)=0P(-2) = 0, giving 2(2)3+3(2)2p(2)+q=02(-2)^3 + 3(-2)^2 - p(-2) + q = 0, which simplifies to 2p+q=42p + q = 4. Solving these equations together gives p=2p = 2 and q=0q = 0. Summing them yields p+q=2p + q = 2.

Adım Adım Çözüm

1
Apply the Remainder Theorem for divisor (x1)(x - 1)
P(1)=2(1)3+3(1)2p(1)+q=3    5p+q=3    pq=2P(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)P(x) by (xa)(x - a) leaves a remainder equal to P(a)P(a).
2
Apply the Factor Theorem for factor (x+2)(x + 2)
P(2)=2(2)3+3(2)2p(2)+q=0    16+12+2p+q=0    2p+q=4P(-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 (xa)(x - a) is a factor of P(x)P(x), then P(a)=0P(a) = 0. Here a=2a = -2.
3
Solve the simultaneous linear equations for pp and qq
Adding pq=2p - q = 2 and 2p+q=42p + q = 4 yields 3p=6    p=23p = 6 \implies p = 2. Substituting p=2p = 2 into pq=2p - q = 2 gives q=0q = 0.
Eliminating qq allows direct calculation of pp, followed by back-substitution for qq.
4
Calculate the target expression p+qp + q
p+q=2+0=2p + q = 2 + 0 = 2
Combine the values of pp and qq to obtain the required sum.

Anahtar Kavram

Factor and Remainder Theorems for Polynomials
Soru 100Soru

If the matrix A=(1022k1314)A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & k & 1 \\ 3 & 1 & 4 \end{pmatrix} is singular, what is the value of kk?

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Cevap: 32\frac{3}{2}

Cevap

32\frac{3}{2}
A matrix is singular if its determinant equals zero. Expanding the determinant of matrix AA along its first row gives 1(4k1)0+2(23k)=2k+31(4k - 1) - 0 + 2(2 - 3k) = -2k + 3. Setting 2k+3=0-2k + 3 = 0 yields k=32k = \frac{3}{2}.

Adım Adım Çözüm

1
Apply the condition for a singular matrix
A matrix is singular when its determinant equals zero, so det(A)=0\det(A) = 0.
By definition, square matrices with zero determinant are singular.
2
Expand det(A)\det(A) along the first row
det(A)=1(4k1)0(83)+2(23k)=4k1+46k=2k+3\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 kk
2k+3=0    2k=3    k=32-2k + 3 = 0 \implies 2k = 3 \implies k = \frac{3}{2}.
Isolating the variable gives the required value of kk.

Anahtar Kavram

Singular Matrix Condition and 3x3 Determinant Expansion
ÖncekiSayfa 5 / 12Sonraki
Algebra Alıştırma Soruları — JAMB UTME — Sayfa 5 | Examkin