Indices and Logarithms Alıştırma Soruları — JAMB UTME

Soru 1Soru →

Solve for $x$ in the logarithmic equation $\log_2 (x + 2) + \log_2 (x - 4) = 4$ .

4

5

6

9

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of

x

is 6.

Using the product rule for logarithms,

\log_2 (x + 2) + \log_2 (x - 4) = \log_2 [(x + 2)(x - 4)] = 4

. Converting to exponential form yields

(x + 2)(x - 4) = 2^4 = 16

. Expanding gives

x^2 - 2x - 8 = 16

, which simplifies to

x^2 - 2x - 24 = 0

. Factoring gives

(x - 6)(x + 4) = 0

. Since the logarithmic arguments must be positive (

x > 4

), the negative root

-4

is discarded, leaving

x = 6

.

Adım Adım Çözüm

1

Apply the product rule of logarithms:

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

\log_2 [(x + 2)(x - 4)] = 4

Logarithms with the same base being added combine by multiplying their arguments.

2

Convert the logarithmic equation to its exponential form

(x + 2)(x - 4) = 2^4 = 16

If

\log_b Y = X

, then

Y = b^X

.

3

Expand and rearrange into a standard quadratic equation

x^2 - 2x - 8 = 16 \implies x^2 - 2x - 24 = 0

Expanding

(x+2)(x-4)

yields

x^2 - 2x - 8

, and subtracting 16 sets the equation to zero.

4

Factor the quadratic equation to find potential solutions

(x - 6)(x + 4) = 0 \implies x = 6 \text{ or } x = -4

The roots of

x^2 - 2x - 24 = 0

are

x = 6

and

x = -4

.

5

Check the domain restrictions for logarithmic functions

x = 6

The arguments of the original logarithms require

x + 2 > 0 \implies x > -2

and

x - 4 > 0 \implies x > 4

. Therefore,

x = -4

is extraneous and

x = 6

is the only valid solution.

Anahtar Kavram

Solving logarithmic equations using product rule and domain constraints

Soru 2Soru →

If $\log_4 x + \log_2 x = 6$ , find the value of $x$ .

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The value of

x

is 16.

Using the change of base identity

\log_b a = \frac{\log_c a}{\log_c b}

, we convert

\log_4 x

into

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

. Substituting this into the equation yields

\frac{3}{2} \log_2 x = 6

, which simplifies to

\log_2 x = 4

, leading directly to

x = 2^4 = 16

.

Adım Adım Çözüm

1

Apply the change of base formula to

\log_4 x

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

Converting all terms to a common base (base 2) simplifies addition of logarithmic terms.

2

Substitute the expression back into the original equation and collect like terms

\frac{1}{2}\log_2 x + \log_2 x = \frac{3}{2}\log_2 x = 6

Adding the coefficients of

\log_2 x

gives

\frac{3}{2}

.

3

Isolate

\log_2 x

\log_2 x = 6 \cdot \frac{2}{3} = 4

Multiplying both sides by

\frac{2}{3}

isolates the logarithmic term.

4

Convert from logarithmic form to exponential form

x = 2^4 = 16

If

\log_b a = c

, then

a = b^c

.

Anahtar Kavram

Change of base rule for logarithms

Soru 3Soru →

What is the positive integer solution to the logarithmic equation $\log_2 x - 3\log_x 2 = 2$ ?

8

6

3

12

Cevabı ve açıklamayı göster

Cevap: $8$

Cevap

The positive integer solution is

8

.

Using the change of base identity

\log_x 2 = \frac{1}{\log_2 x}

, we rewrite the equation as

u - \frac{3}{u} = 2

where

u = \log_2 x

. Rearranging yields the quadratic

u^2 - 2u - 3 = 0

, which factors as

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

. This gives

u = 3

or

u = -1

. Converting back to

x = 2^u

, we get

x = 2^3 = 8

or

x = 2^{-1} = \frac{1}{2}

. Since the question asks for the positive integer solution, the correct answer is 8.

Adım Adım Çözüm

1

Apply the change of base rule to express

\log_x 2

in terms of base 2.

\log_2 x - \frac{3}{\log_2 x} = 2

The change of base identity states that

\log_b a = \frac{1}{\log_a b}

.

2

Substitute

u = \log_2 x

into the equation and clear the fraction.

u - \frac{3}{u} = 2 \implies u^2 - 2u - 3 = 0

Multiplying through by

u

(where

u \neq 0

) transforms the equation into standard quadratic form.

3

Factor the quadratic equation to solve for

u

.

(u - 3)(u + 1) = 0 \implies u = 3 \text{ or } u = -1

Finding the roots of the quadratic equation in terms of

u

.

4

Convert back to

x

using

x = 2^u

and select the positive integer solution.

For

u = 3

:

x = 2^3 = 8

. For

u = -1

:

x = 2^{-1} = \frac{1}{2}

.

The question asks specifically for the positive integer solution, which is 8.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction

Soru 4Soru →

Find the value of $x$ that satisfies the exponential equation $27^{x - 1} = \frac{9^{x + 1}}{3^{x - 5}}$ .

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Cevap: 5

Cevap

The value of

x

is 5.

Rewriting the terms with a common base of 3 transforms the equation into

3^{3x-3} = 3^{x+7}

. Equating exponents gives

3x - 3 = x + 7

, which yields

x = 5

.

Adım Adım Çözüm

1

Convert all terms to base 3

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

and

9^{x+1} = 3^{2x+2}

Laws of indices require identical bases to manipulate exponents.

2

Apply division rule of indices to the right-hand side

\frac{3^{2x+2}}{3^{x-5}} = 3^{x+7}

When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Equate the exponents and solve for

x

3x - 3 = x + 7 \implies 2x = 10 \implies x = 5

If

a^f(x) = a^g(x)

for

a > 0

and

a \neq 1

, then

f(x) = g(x)

.

Anahtar Kavram

Solving exponential equations using base reduction and exponent laws

Tahmini Süre:1m 30s

Soru 5Soru →

If $2^{x+y} = 32$ and $\log_3 x + \log_3 y = 1 + \log_3 2$ , what is the value of $x^2 + y^2$ ?

13

21

26

37

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Cevap: $13$

Cevap

The value of

x^2 + y^2

is

13

.

Expressing

32

as

2^5

gives

x + y = 5

. Converting

1

to

\log_3 3

allows combining the right-hand side to

\log_3(3 \times 2) = \log_3 6

, so

\log_3(xy) = \log_3 6 \implies xy = 6

. Evaluating

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

yields

5^2 - 2(6) = 25 - 12 = 13

.

Adım Adım Çözüm

1

Simplify the exponential equation using index laws.

x + y = 5

Since

32 = 2^5

,

2^{x+y} = 2^5 \implies x + y = 5

.

2

Simplify the logarithmic equation using logarithm laws.

xy = 6

\log_3 x + \log_3 y = \log_3(xy)

and

1 + \log_3 2 = \log_3 3 + \log_3 2 = \log_3(3 \times 2) = \log_3 6

. Therefore,

xy = 6

.

3

Use the algebraic identity to find

x^2 + y^2

.

x^2 + y^2 = 13

Using

(x+y)^2 = x^2 + y^2 + 2xy

, we have

5^2 = x^2 + y^2 + 2(6) \implies 25 = x^2 + y^2 + 12 \implies x^2 + y^2 = 13

.

Anahtar Kavram

Solving simultaneous equations involving indices and logarithms using index laws and logarithmic identities.

Soru 6Soru →

If $\log_{10}(x) + \log_{10}(5) = 2$ , what is the value of $x$ ?

20

95

10

40

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

20

Using the logarithmic product identity

\log_b(A) + \log_b(B) = \log_b(AB)

, the left side simplifies to

\log_{10}(5x) = 2

. Converting to exponential form gives

5x = 10^2 = 100

, which simplifies directly to

x = 20

.

Adım Adım Çözüm

1

Apply the product law of logarithms

\log_{10}(5x) = 2

The sum of logarithms with the same base equals the logarithm of their product:

\log_b(A) + \log_b(B) = \log_b(AB)

.

2

Convert the logarithmic equation to exponential form

5x = 10^2 = 100

By definition of logarithms, if

\log_b(y) = c

, then

y = b^c

.

3

Solve for

x

x = \frac{100}{5} = 20

Divide both sides of the linear equation by

5

.

Anahtar Kavram

Logarithmic Product Law and Exponential Conversion

Tahmini Süre:45s

Soru 7Soru →

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

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Simultaneous Exponential and Logarithmic Systems

Tahmini Süre:2m 0s

Soru 8Soru →

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

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving exponential equations reducible to quadratic form using laws of indices

Soru 9Soru →

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

20

12

48

3

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Change of base rule and solving logarithmic quadratic equations

Soru 10Soru →

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

2

\frac{8}{3}

5

\frac{2}{3}

Cevabı ve açıklamayı göster

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

Cevap

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}

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Solving Exponential Equations using Base Conversion

Tahmini Süre:1m 0s

Soru 11Soru →

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

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

Soru 12Soru →

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

27

9

3

81

Cevabı ve açıklamayı göster

Cevap: $27$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Change of base formula for logarithms:

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

Tahmini Süre:1m 30s

Soru 13Soru →

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

Cevabı ve açıklamayı göster

Cevap: 81

Cevap

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.

Adım Adım Çözüm

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}

.

Anahtar Kavram

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

Soru 14Soru →

Find the value of $x$ that satisfies the logarithmic equation $\log_2(x^2 - 1) - \log_2(x - 1) = 3$ .

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

x

is

7

.

Applying the logarithm quotient law reduces

\log_2(x^2 - 1) - \log_2(x - 1)

to

\log_2\left(\frac{x^2 - 1}{x - 1}\right)

. Factoring

x^2 - 1

into

(x - 1)(x + 1)

and canceling

(x - 1)

leaves

\log_2(x + 1) = 3

. Converting this to exponential form gives

x + 1 = 2^3 = 8

, which yields

x = 7

.

Adım Adım Çözüm

1

Apply the quotient rule for logarithms to combine the left-hand side into a single logarithmic expression.

\log_2\left(\frac{x^2 - 1}{x - 1}\right) = 3

According to logarithmic rules, the difference of two logarithms of the same base is equal to the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Factor the numerator as a difference of two squares and simplify the rational expression.

\log_2(x + 1) = 3

Since

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

, dividing by

(x - 1)

leaves

(x + 1)

, given

x \neq 1

.

3

Rewrite the logarithmic equation in exponential form.

x + 1 = 2^3 = 8

By definition of logarithms,

\log_b y = z \iff b^z = y

.

4

Solve the resulting linear equation for

x

.

x = 7

Subtracting

1

from both sides gives

x = 7

.

Anahtar Kavram

Quotient Law of Logarithms and Logarithmic Equations

Tahmini Süre:1m 30s

Soru 15Soru →

Find the real value of $x$ that satisfies the equation $\log_2 (2^{2x} + 12) - x = 3 - \log_2 (2^x - 1)$ .

1

2

3

0

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Cevap: 1

Cevap

1

By bringing all terms involving

x

together, the equation becomes

\log_2(2^{2x}+12) + \log_2(2^x-1) - \log_2(2^x) = 3

. Combining these logarithmic terms yields

\log_2\left(\frac{(2^{2x}+12)(2^x-1)}{2^x}\right) = 3

. Converting to exponential form gives

\frac{(2^{2x}+12)(2^x-1)}{2^x} = 8

. Setting

u = 2^x

converts this into the cubic equation

u^3 - u^2 + 4u - 12 = 0

. Factoring out

(u - 2)

leaves

(u - 2)(u^2 + u + 6) = 0

. Since

u^2 + u + 6 = 0

has no real solutions,

u = 2

is the only real root. Hence

2^x = 2

, which gives

x = 1

.

Adım Adım Çözüm

1

Rearrange the equation to group all logarithmic terms on one side.

\log_2 (2^{2x} + 12) + \log_2 (2^x - 1) - x = 3

Grouping logarithmic terms allows the application of logarithm combination laws.

2

Express

x

as

\log_2 (2^x)

and combine the logarithms using log laws.

\log_2 \left( \frac{(2^{2x} + 12)(2^x - 1)}{2^x} \right) = 3

Product and quotient laws of logarithms allow single logarithmic expression formation:

\log_2 A + \log_2 B - \log_2 C = \log_2\left(\frac{AB}{C}\right)

.

3

Convert from logarithmic to exponential form.

\frac{(2^{2x} + 12)(2^x - 1)}{2^x} = 2^3 = 8

Definition of logarithm:

\log_b(Y) = Z \implies Y = b^Z

.

4

Substitute

u = 2^x

(where

u > 1

) to form an algebraic equation.

\frac{(u^2 + 12)(u - 1)}{u} = 8 \implies (u^2 + 12)(u - 1) = 8u

Since

2^{2x} = (2^x)^2 = u^2

, substitution simplifies the exponential expression into a polynomial.

5

Expand and simplify into a standard cubic equation.

u^3 - u^2 + 12u - 12 = 8u \implies u^3 - u^2 + 4u - 12 = 0

Standard polynomial form enables root searching via polynomial division and factoring.

6

Factor the cubic equation.

(u - 2)(u^2 + u + 6) = 0

Testing

u = 2

gives

2^3 - 2^2 + 4(2) - 12 = 8 - 4 + 8 - 12 = 0

, confirming

(u - 2)

is a factor.

7

Solve for real values of

u

and

x

.

u = 2 \implies 2^x = 2 \implies x = 1

The quadratic factor

u^2 + u + 6 = 0

has discriminant

\Delta = 1 - 24 = -23 < 0

(no real roots). Thus

u = 2

is the unique solution.

Anahtar Kavram

Logarithmic equations requiring exponential substitution and polynomial factorization

Tahmini Süre:2m 30s

Soru 16Soru →

If $\log_2 (x + 3) + \log_2 4 = 5$ , what is the value of $x$ ?

25

5

18

3

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Cevap: $5$

Cevap

The value of

x

is

5

.

Using the logarithmic addition property

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

, the equation

\log_2(x+3) + \log_2 4 = 5

becomes

\log_2[4(x+3)] = 5

. Converting to index form gives

4(x+3) = 2^5 = 32

. Dividing by

4

yields

x + 3 = 8

, so

x = 5

.

Adım Adım Çözüm

1

Apply the product rule of logarithms:

\log_b A + \log_b B = \log_b (A \cdot B)

\log_2 [4(x + 3)] = 5

Logarithms with the same base that are added together can be combined by multiplying their arguments.

2

Convert the logarithmic equation into its equivalent exponential form

4(x + 3) = 2^5 = 32

If

\log_b Y = c

, then

Y = b^c

.

3

Divide both sides by 4 and solve for

x

x + 3 = 8 \implies x = 5

Linear algebraic simplification.

Anahtar Kavram

Product Law of Logarithms and Log-to-Exponential Conversion

Soru 17Soru →

What is the sum of all real values of $x$ that satisfy the logarithmic equation $\log_3(9^x + 81) - x = \log_3 30$ ?

4

10

30

6

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The sum of all real values of

x

satisfying the equation is 4.

Using logarithmic properties,

\log_3(9^x + 81) - x = \log_3 30

simplifies to

\frac{9^x + 81}{3^x} = 30

. Substituting

u = 3^x

produces the quadratic equation

u^2 - 30u + 81 = 0

, which yields

u = 3

and

u = 27

. Solving

3^x = 3

and

3^x = 27

gives

x = 1

and

x = 3

. Their sum is

1 + 3 = 4

.

Adım Adım Çözüm

1

Express

x

as

\log_3(3^x)

and apply the logarithm quotient rule.

\log_3\left(\frac{9^x + 81}{3^x}\right) = \log_3 30

Using

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

combines terms into a single logarithm.

2

Equate the logarithmic arguments and split the fraction.

\frac{9^x + 81}{3^x} = 30 \implies 3^x + \frac{81}{3^x} = 30

Logarithmic equality with identical bases implies equality of their arguments.

3

Substitute

u = 3^x

to form and solve a quadratic equation.

u + \frac{81}{u} = 30 \implies u^2 - 30u + 81 = 0 \implies (u - 3)(u - 27) = 0 \implies u = 3 \text{ or } u = 27

Since

9^x = (3^x)^2 = u^2

, substituting

u = 3^x

converts the equation into a standard quadratic equation.

4

Solve for

x

from

3^x = u

and calculate the sum of the solutions.

3^x = 3 \implies x = 1; \quad 3^x = 27 \implies x = 3. \quad \text{Sum} = 1 + 3 = 4

Matching exponential bases gives

x = 1

and

x = 3

.

Anahtar Kavram

Logarithmic quotient identity and quadratic substitution for exponential equations

Tahmini Süre:2m 0s

Soru 18Soru →

If $\log_{10} 2 = p$ and $\log_{10} 3 = q$ , which of the following expressions represents $\log_{10} 18$ in terms of $p$ and $q$ ?

p + 2q

2pq

2p + q

p + q^2

Cevabı ve açıklamayı göster

Cevap: $p + 2q$

Cevap

The expression for

\log_{10} 18

in terms of

p

and

q

is

p + 2q

.

Expanding

\log_{10} 18

as

\log_{10}(2 \times 3^2)

gives

\log_{10} 2 + 2\log_{10} 3

, which simplifies directly to

p + 2q

.

Adım Adım Çözüm

1

Factorize 18 into prime factors.

18 = 2 \times 3^2

Decomposing 18 into prime factors allows the application of the given logarithm values for 2 and 3.

2

Apply the product rule of logarithms:

\log_{10}(a \times b) = \log_{10} a + \log_{10} b

.

\log_{10} 18 = \log_{10}(2 \times 3^2) = \log_{10} 2 + \log_{10}(3^2)

The logarithm of a product equals the sum of the logarithms of individual factors.

3

Apply the power rule of logarithms:

\log_{10}(b^n) = n \log_{10} b

.

\log_{10}(3^2) = 2 \log_{10} 3

The logarithm of a power expression allows bringing the exponent to the front as a multiplier.

4

Substitute the defined values

p = \log_{10} 2

and

q = \log_{10} 3

.

\log_{10} 18 = p + 2q

Replaces logarithmic terms with their algebraic representations.

Anahtar Kavram

Logarithm Expansion Laws (Product and Power Rules)

Soru 19Soru →

Find the real value of $x$ that satisfies the logarithmic equation $\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The real value of

x

that satisfies the equation is

6

.

Applying the logarithmic quotient rule transforms

\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3

into

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

. Expressing this in exponential form yields

\frac{x^2 + 3x - 22}{x - 2} = 8

, which simplifies to

x^2 - 5x - 6 = 0

. Factoring gives

(x - 6)(x + 1) = 0

. Since the argument of a logarithm must be strictly positive (

x - 2 > 0 \implies x > 2

),

x = -1

is extraneous and

x = 6

is the only valid solution.

Adım Adım Çözüm

1

Combine the logarithmic terms using the quotient law of logarithms.

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

The difference of two logarithms with the same base equals the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Convert the logarithmic equation into its equivalent exponential form.

\frac{x^2 + 3x - 22}{x - 2} = 2^3 = 8

If

\log_b Y = c

, then

Y = b^c

.

3

Clear the denominator and simplify to form a quadratic equation.

x^2 - 5x - 6 = 0

Multiplying both sides by

(x - 2)

yields

x^2 + 3x - 22 = 8x - 16

, which rearranges to

x^2 - 5x - 6 = 0

.

4

Solve the quadratic equation by factoring.

(x - 6)(x + 1) = 0 \implies x = 6 \text{ or } x = -1

The factors of

-6

that sum to

-5

are

-6

and

1

.

5

Test roots against domain restrictions to eliminate extraneous solutions.

x = 6

Logarithmic arguments must be strictly positive. For

\log_2(x - 2)

to be defined,

x > 2

. Thus,

x = -1

is extraneous, leaving

x = 6

as the unique valid solution.

Anahtar Kavram

Logarithmic Equations and Domain Restrictions

Soru 20Soru →

Find the real value of $x$ that satisfies the logarithmic equation $\log_5(x^2 - 4) - \log_5(x - 2) = 2$ .

Cevabı ve açıklamayı göster

Cevap: 23

Cevap

The value of

x

that satisfies the equation is 23.

Applying the logarithmic quotient rule gives

\log_5 \left(\frac{x^2 - 4}{x - 2}\right) = 2

. Factoring the numerator gives

\log_5 \left(\frac{(x - 2)(x + 2)}{x - 2}\right) = \log_5(x + 2) = 2

. Converting to exponential form yields

x + 2 = 5^2 = 25

, which simplifies to

x = 23

.

Adım Adım Çözüm

1

Apply the logarithmic quotient rule

\log_5\left(\frac{x^2 - 4}{x - 2}\right) = 2

The difference of two logarithms of the same base equals the logarithm of their quotient.

2

Factor the numerator and simplify the expression

\log_5(x + 2) = 2

Factoring

x^2 - 4

into

(x-2)(x+2)

allows canceling the common factor

(x-2)

in the denominator.

3

Convert the logarithmic equation into exponential form

x + 2 = 5^2 = 25

By definition,

\log_b(y) = c

is equivalent to

b^c = y

.

4

Solve the linear equation for

x

x = 23

Subtracting 2 from both sides gives

x = 23

.

Anahtar Kavram

Logarithmic Quotient Rule and Logarithm-to-Exponent Conversion

Indices and Logarithms

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