Indices and Laws of Indices Practice Questions — JAMB UTME

Question 1Question →

Find the real value of $x$ that satisfies the exponential equation $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$ .

Show answer & explanation

Answer: 1.5

Answer

The value of

x

is

1.5

(or

\frac{3}{2}

).

By using index laws to group base-2 terms on the left side and base-3 terms on the right side, we obtain

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

. Rearranging gives

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}

. Equating indices gives

x = 1.5

.

Step-by-Step Solution

1

Group like exponential terms with base 2 and base 3 on opposite sides of the equation.

4^x + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

Grouping terms with common prime bases allows for factoring exponential terms.

2

Apply the product and power laws of indices:

4^x = 2^{2x}

,

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

,

3^{x \pm \frac{1}{2}} = 3^x \cdot 3^{\pm \frac{1}{2}}

.

2^{2x} + \frac{1}{2} \cdot 2^{2x} = 3^x \cdot \sqrt{3} + 3^x \cdot \frac{1}{\sqrt{3}}

Separating the variable exponents from constant exponents prepares each side for factoring.

3

Factor out

2^{2x}

from the left side and

3^x

from the right side, then simplify arithmetic terms.

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

Factoring isolates the variable terms

2^{2x}

and

3^x

from numerical constants.

4

Divide to form the ratio

\frac{4^x}{3^x} = \left(\frac{4}{3}\right)^x

and simplify the numerical fraction on the right.

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}}

Expressing both sides with unified variable bases facilitates solving for

x

by equating powers.

5

Rewrite

\frac{8}{3\sqrt{3}}

as a power of

\frac{4}{3}

and solve for

x

.

\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{\frac{3}{2}} \implies x = \frac{3}{2} = 1.5

Since

\frac{8}{3\sqrt{3}} = \frac{4^{3/2}}{3^{3/2}} = (\frac{4}{3})^{3/2}

, equating exponents yields

x = 1.5

.

Key Concept

Solving mixed-base exponential equations by grouping, factoring, and converting to a unified base ratio.

Estimated Time:3m 0s

Question 2Question →

If $2^{2x + 1} - 9(2^x) + 4 = 0$ , what is the product of all real values of $x$ that satisfy the equation?

-2

2

1

-1

Show answer & explanation

Answer: $-2$

Answer

The product of all real values of

x

satisfying the equation is

-2

.

Rewriting

2^{2x+1}

as

2(2^x)^2

allows substitution of

u = 2^x

, giving

2u^2 - 9u + 4 = 0

. Solving for

u

gives

u = \frac{1}{2}

and

u = 4

. Converting back to

x

via

2^x = 2^{-1}

and

2^x = 2^2

yields

x = -1

and

x = 2

. The product of these roots is

(-1) \times 2 = -2

.

Step-by-Step Solution

1

Apply the law of indices to rewrite the first term.

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

. Thus, the equation becomes

2(2^x)^2 - 9(2^x) + 4 = 0

.

Splitting the index using

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

enables transformation into a quadratic form.

2

Substitute

u = 2^x

and solve the resulting quadratic equation.

2u^2 - 9u + 4 = 0 \implies (2u - 1)(u - 4) = 0 \implies u = \frac{1}{2} \text{ or } u = 4

.

Algebraic substitution simplifies the exponential equation into a standard quadratic equation.

3

Convert the values of

u

back to

x

using index laws.

For

u = \frac{1}{2}

,

2^x = 2^{-1} \implies x_1 = -1

. For

u = 4

,

2^x = 2^2 \implies x_2 = 2

.

Equating bases (

2^x = 2^k \implies x = k

) isolates the unknown variable

x

.

4

Compute the product of the solutions

x_1

and

x_2

.

Product =

x_1 \cdot x_2 = (-1) \times 2 = -2

.

The question specifically requests the product of all real values of

x

.

Key Concept

Solving quadratic-form exponential equations using index laws and variable substitution.

Estimated Time:2m 0s

Question 3Question →

Find the real value of

x

that satisfies the exponential equation

\frac{8^{x + 2}}{4^{x - 1}} = 16^{x - 1}

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

Rewriting the terms in base 2 gives

2^{3(x+2)} / 2^{2(x-1)} = 2^{4(x-1)}

. Applying the quotient rule gives an exponent of

(3x + 6) - (2x - 2) = x + 8

on the left. Equating the exponents yields

x + 8 = 4x - 4

, which solves cleanly to

x = 4

.

Step-by-Step Solution

1

Express all terms using a common prime base of 2

The equation becomes

\frac{(2^3)^{x + 2}}{(2^2)^{x - 1}} = (2^4)^{x - 1}

.

Converting to a common base enables the use of index laws to simplify the equation.

2

Apply the power-of-a-power and quotient laws of indices

The left side simplifies to

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

and the right side is

2^{4x - 4}

.

When dividing powers of the same base, exponents are subtracted:

a^m \div a^n = a^{m-n}

.

3

Equate exponents and solve the linear equation

x + 8 = 4x - 4 \implies 3x = 12 \implies x = 4

.

Because the bases on both sides are identical, their respective exponents must be equal.

Key Concept

Solving exponential equations using common base conversion and laws of indices

Question 4Question →

What is the simplified form of the expression $\frac{a^3 \times a^4}{a^2}$ ?

a^5

a^{10}

a^6

a^9

Show answer & explanation

Answer: $a^5$

Answer

The simplified expression is

a^5

.

Combining terms with the same base using the product law gives

a^{3+4} = a^7

in the numerator. Next, dividing by

a^2

using the quotient law yields

a^{7-2} = a^5

.

Step-by-Step Solution

1

Simplify the numerator using the product law of indices

a^3 \times a^4 = a^{3+4} = a^7

When multiplying terms with the same base, add their exponents.

2

Divide by the denominator using the quotient law of indices

\frac{a^7}{a^2} = a^{7-2} = a^5

When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.

Key Concept

Laws of Indices (Product and Quotient Rules)

Estimated Time:45s

Question 5Question →

If $x$ , $y$ , and $z$ are non-zero real numbers satisfying the exponential equation $2^x = 5^y = 100^z$ , what is the numerical value of the expression $z \left( \frac{2}{x} + \frac{2}{y} \right)$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of

z \left( \frac{2}{x} + \frac{2}{y} \right)

is

1

.

By setting

2^x = 5^y = 100^z = k

, we can write

2 = k^{1/x}

,

5 = k^{1/y}

, and

100 = k^{1/z}

. Factoring

100 = 2^2 \times 5^2

gives

k^{1/z} = (k^{1/x})^2 \times (k^{1/y})^2 = k^{2/x + 2/y}

. Equating exponents gives

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

, which upon multiplying by

z

yields

1

.

Step-by-Step Solution

1

Equate the given exponential expressions to a common constant

k

.

2^x = 5^y = 100^z = k

Introducing a common variable allows isolating each base exponent combination.

2

Express bases

2

,

5

, and

100

in terms of

k

using fractional indices.

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

,

5 = k^{\frac{1}{y}}

,

100 = k^{\frac{1}{z}}

Applying the power law

(a^m)^{\frac{1}{m}} = a

isolates each base.

3

Express

100

using prime factorization of the other bases.

100 = 2^2 \times 5^2

Establishing a numerical relationship between

100

,

2

, and

5

links the exponential variables.

4

Substitute the

k

-expressions into

100 = 2^2 \times 5^2

and apply index multiplication laws.

k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^2 \times \left(k^{\frac{1}{y}}\right)^2 = k^{\frac{2}{x}} \times k^{\frac{2}{y}} = k^{\frac{2}{x} + \frac{2}{y}}

Multiplying powers with the same base requires adding the exponents:

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

.

5

Equate exponents of identical bases.

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

If

k^a = k^b

for

k > 1

, then

a = b

.

6

Multiply both sides of the equation by

z

.

z \left( \frac{2}{x} + \frac{2}{y} \right) = 1

Rearranging the equation yields the exact numerical value of the requested expression.

Key Concept

Equating Exponents of Common Bases and Fractional Indices

Question 6Question →

What is the value of

x

that satisfies the exponential equation

\frac{4^{2x + 1} \times 8^{x - 1}}{16^x} = 32

?

2

1

\frac{5}{3}

11

Show answer & explanation

Answer: $2$

Answer

The value of

x

is

2

.

Converting all terms to base

2

gives

\frac{2^{4x+2} \times 2^{3x-3}}{2^{4x}} = 2^5

. Combining the powers on the left side yields

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

. Equating exponents gives

3x - 1 = 5

, which solves to

x = 2

.

Step-by-Step Solution

1

Express all base numbers in terms of a common prime base (base 2).

4 = 2^2

,

8 = 2^3

,

16 = 2^4

, and

32 = 2^5

.

Converting all terms to powers of 2 enables the application of index laws.

2

Substitute these prime base powers into the original equation and expand exponents.

\frac{(2^2)^{2x + 1} \times (2^3)^{x - 1}}{(2^4)^x} = 2^5 \implies \frac{2^{4x + 2} \times 2^{3x - 3}}{2^{4x}} = 2^5

Applying

(a^m)^n = a^{m \cdot n}

requires multiplying the outer exponent by every term in the inner exponent.

3

Apply index laws for multiplication (

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

) and division (

a^m \div a^n = a^{m-n}

) on the left-hand side.

2^{(4x + 2) + (3x - 3) - 4x} = 2^5 \implies 2^{3x - 1} = 2^5

Powers with the same base are combined by adding numerator exponents and subtracting denominator exponents.

4

Equate the exponents since the bases are identical, and solve for

x

.

3x - 1 = 5 \implies 3x = 6 \implies x = 2

If

a^x = a^y

for

a > 0

and

a \neq 1

, then

x = y

.

Key Concept

Solving Exponential Equations using Laws of Indices

Estimated Time:1m 30s

Question 7Question →

Find the real value of

x

that satisfies the exponential equation

5^{2x - 1} \times 25^{x + 1} = 125^{x + 2}

Show answer & explanation

Answer: 5

Answer

The value of

x

is 5.

By converting all terms to base 5 (

25 = 5^2

and

125 = 5^3

), the equation becomes

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

. Simplifying exponents gives

5^{2x-1} \times 5^{2x+2} = 5^{3x+6}

. Adding the left-hand powers results in

5^{4x+1} = 5^{3x+6}

. Equating the exponents yields

4x + 1 = 3x + 6

, leading directly to

x = 5

.

Step-by-Step Solution

1

Express all terms with a common base of 5

5^{2x-1} \times 5^{2x+2} = 5^{3x+6}

Since

25 = 5^2

and

125 = 5^3

, using index laws

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

allows all expressions to share base 5.

2

Apply the product rule of indices on the left side

5^{4x+1} = 5^{3x+6}

According to the product law

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

, the powers are added:

(2x-1) + (2x+2) = 4x+1

.

3

Equate powers of equal bases to solve for x

x = 5

Since bases are equal, exponents must be equal:

4x + 1 = 3x + 6 \implies 4x - 3x = 6 - 1 \implies x = 5

.

Key Concept

Indices and Laws of Indices

Question 8Question →

If $4^{2x + 1} \times 8^{1 - x} = 32^{x - 1}$ , find the value of $x$ .

\frac{5}{2}

\frac{9}{4}

\frac{3}{2}

3

Show answer & explanation

Answer: $\frac{5}{2}$

Answer

The value of

x

is

\frac{5}{2}

.

Converting all terms to base

2

yields

4^{2x+1} = 2^{4x+2}

,

8^{1-x} = 2^{3-3x}

, and

32^{x-1} = 2^{5x-5}

. Applying the law

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

gives

2^{(4x+2)+(3-3x)} = 2^{x+5}

. Equating exponents gives

x+5 = 5x-5

, which solves cleanly to

x = \frac{5}{2}

.

Step-by-Step Solution

1

Express all terms with a common prime base (base 2)

4 = 2^2

,

8 = 2^3

, and

32 = 2^5

To apply the laws of indices, all terms must share the same base.

2

Apply the power of a power law

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

to each term

(2^2)^{2x+1} \times (2^3)^{1-x} = (2^5)^{x-1} \implies 2^{2(2x+1)} \times 2^{3(1-x)} = 2^{5(x-1)} \implies 2^{4x+2} \times 2^{3-3x} = 2^{5x-5}

Multiplying the inner exponent by the outer exponent simplifies nested powers.

3

Apply the multiplication law

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

on the left-hand side

2^{(4x+2) + (3-3x)} = 2^{5x-5} \implies 2^{x+5} = 2^{5x-5}

Powers with the same base being multiplied require adding their exponents.

4

Equate the exponents and solve for

x

x + 5 = 5x - 5 \implies 5 + 5 = 5x - x \implies 10 = 4x \implies x = \frac{10}{4} = \frac{5}{2}

Since the bases are equal and non-zero, their exponents must be equal.

Key Concept

Solving exponential equations by expressing numbers in terms of a common prime base and applying laws of indices.

Question 9Question →

Find the value of

x

that satisfies the exponential equation

\frac{4^{x+1} + 4^x + 4^{x-1}}{2^{x+2} + 2^{x+1} + 2^x} = 24

Show answer & explanation

Answer: 5

Answer

5

Factoring

4^{x-1}

from the numerator yields

4^{x-1}(16 + 4 + 1) = 21 \cdot 4^{x-1}

. Factoring

2^x

from the denominator yields

2^x(4 + 2 + 1) = 7 \cdot 2^x

. Dividing the numerical coefficients gives

\frac{21}{7} = 3

. Substituting

4^{x-1} = 2^{2x-2}

into the ratio gives

3 \cdot \frac{2^{2x-2}}{2^x} = 3 \cdot 2^{x-2}

. Setting

3 \cdot 2^{x-2} = 24

leads to

2^{x-2} = 8 = 2^3

, which gives

x - 2 = 3

and therefore

x = 5

.

Step-by-Step Solution

1

Factor out common terms from the numerator and denominator

Numerator:

4^{x-1}(4^2 + 4^1 + 1) = 21 \cdot 4^{x-1}

. Denominator:

2^x(2^2 + 2^1 + 1) = 7 \cdot 2^x

.

Grouping power terms simplifies sums of exponential expressions.

2

Divide the numerator by the denominator and convert bases

\frac{21 \cdot 4^{x-1}}{7 \cdot 2^x} = 3 \cdot \frac{(2^2)^{x-1}}{2^x} = 3 \cdot \frac{2^{2x-2}}{2^x}

Simplifying

\frac{21}{7} = 3

and expressing base 4 in base 2 allows applying laws of indices.

3

Apply the quotient rule of indices:

\frac{a^m}{a^n} = a^{m-n}

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

Subtracting exponents of like bases simplifies the fractional index expression.

4

Set the simplified expression equal to 24 and solve for

x

3 \cdot 2^{x-2} = 24 \implies 2^{x-2} = 8 = 2^3 \implies x - 2 = 3 \implies x = 5

Equating the exponents when bases are identical yields the linear equation

x - 2 = 3

.

Key Concept

Factoring sums of exponential terms and applying the quotient rule of indices

Question 10Question →

Find the value of

x

that satisfies the exponential equation

\sqrt{\frac{125^{x+1}}{5^{x-1}}} = 25^{x-1}

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

Converting all terms to base 5 yields

5^{x+2}

on the left-hand side and

5^{2x-2}

on the right-hand side. Setting the exponents equal gives

x + 2 = 2x - 2

, which solves to

x = 4

.

Step-by-Step Solution

1

Express all terms with a common base of 5

125 = 5^3

and

25 = 5^2

Converting terms to prime base 5 allows the application of standard laws of indices.

2

Simplify the fraction inside the square root

\frac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3) - (x-1)} = 5^{2x+4}

Subtract the denominator exponent from the numerator exponent when dividing like bases.

3

Apply the square root as a fractional exponent

\sqrt{5^{2x+4}} = (5^{2x+4})^{1/2} = 5^{x+2}

Taking the square root of a power is equivalent to multiplying the exponent by 1/2.

4

Equate the simplified exponents of both sides

x + 2 = 2x - 2

With identical bases of 5 on both sides, the exponents must be equal.

5

Solve the linear equation for x

x = 4

Rearranging terms gives

2x - x = 2 + 2

, which yields

x = 4

.

Key Concept

Indices and Laws of Indices

Question 11Question →

Find the value of

x

that satisfies the exponential equation

\frac{100^{x+1}}{1000^{x-1}} = 10^{2x-1}

Show answer & explanation

Answer: 2

Answer

The value of

x

that satisfies the equation is 2.

Rewriting the terms in base 10 gives

\frac{10^{2x+2}}{10^{3x-3}} = 10^{2x-1}

. Using the division rule of indices yields

10^{-x+5} = 10^{2x-1}

. Equating exponents gives

-x+5 = 2x-1

, which simplifies to

3x = 6

, so

x = 2

.

Step-by-Step Solution

1

Convert each power to base 10

100^{x+1} = (10^2)^{x+1} = 10^{2x+2}

and

1000^{x-1} = (10^3)^{x-1} = 10^{3x-3}

Converting all non-prime composite bases to powers of a common fundamental base allows exponent comparison.

2

Apply the quotient rule of indices to the left side

\frac{10^{2x+2}}{10^{3x-3}} = 10^{(2x+2)-(3x-3)} = 10^{-x+5}

According to the index quotient law

\frac{a^m}{a^n} = a^{m-n}

, subtract the denominator's exponent from the numerator's exponent.

3

Equate exponents of equal bases

-x + 5 = 2x - 1

If

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

for

a > 0, a \neq 1

, then

f(x) = g(x)

.

4

Solve the linear equation for

x

3x = 6 \implies x = 2

Isolate the variable

x

to find its value.

Key Concept

Exponential equations solvable by converting to a common base

Question 12Question →

What is the value of $x$ that satisfies the equation $16^{x - 1} \times 4^{x + 2} = 8^{x + 3}$ ?

3

\frac{1}{3}

\frac{8}{3}

\frac{3}{2}

Show answer & explanation

Answer: $3$

Answer

The value of

x

is

3

.

Converting all terms to base 2 gives

2^{4(x-1)} \times 2^{2(x+2)} = 2^{3(x+3)}

. Expanding the exponents gives

2^{4x-4} \times 2^{2x+4} = 2^{3x+9}

, which simplifies to

2^{6x} = 2^{3x+9}

. Equating exponents yields

6x = 3x + 9

, giving the solution

x = 3

.

Step-by-Step Solution

1

Express all terms with a common base of 2.

16 = 2^4

,

4 = 2^2

, and

8 = 2^3

, so the equation becomes

(2^4)^{x-1} \times (2^2)^{x+2} = (2^3)^{x+3}

.

Converting all terms to a common prime base allows application of index laws.

2

Apply power of a power law

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

.

2^{4x - 4} \times 2^{2x + 4} = 2^{3x + 9}

.

Multiply exponents when raising a power to another power.

3

Apply multiplication law

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

on the left side.

2^{(4x - 4) + (2x + 4)} = 2^{6x}

.

Add exponents when multiplying powers with the same base.

4

Equate exponents of equal bases.

6x = 3x + 9 \implies 3x = 9 \implies x = 3

.

If

a^m = a^n

for

a > 0

and

a \neq 1

, then

m = n

.

Key Concept

Laws of Indices and Solving Exponential Equations

Question 13Question →

Given that

x

and

y

are real numbers satisfying the simultaneous exponential equations

3^x \times 9^y = 81

and

8^x \times 4^y = 256

find the value of

x^2 + y^2

.

Show answer & explanation

Answer: 5

Answer

The value of

x^2 + y^2

is 5.

Converting all terms to their prime bases (

3

for the first equation and

2

for the second equation) produces the simultaneous linear equations

x + 2y = 4

and

3x + 2y = 8

. Solving these gives

x = 2

and

y = 1

. Substituting these values into

x^2 + y^2

gives

2^2 + 1^2 = 5

.

Step-by-Step Solution

1

Convert all terms in the first equation to powers of base 3.

x + 2y = 4

Since

9 = 3^2

and

81 = 3^4

, applying the product law of indices

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

gives

3^{x+2y} = 3^4

. Equating the exponents gives

x + 2y = 4

.

2

Convert all terms in the second equation to powers of base 2.

3x + 2y = 8

Since

8 = 2^3

,

4 = 2^2

, and

256 = 2^8

, applying the laws of indices yields

2^{3x} \times 2^{2y} = 2^8 \implies 2^{3x+2y} = 2^8

. Equating exponents gives

3x + 2y = 8

.

3

Solve the system of simultaneous linear equations for

x

and

y

.

x = 2

and

y = 1

Subtracting

x + 2y = 4

from

3x + 2y = 8

yields

2x = 4 \implies x = 2

. Substituting

x = 2

into

x + 2y = 4

gives

2 + 2y = 4 \implies y = 1

.

4

Evaluate the target expression

x^2 + y^2

.

5

Substitute

x = 2

and

y = 1

into

x^2 + y^2

to obtain

2^2 + 1^2 = 4 + 1 = 5

.

Key Concept

Converting exponential terms to common prime bases to reduce exponential equations into linear equations.

Question 14Question →

If $27^{x - 1} = 9^{x + 1}$ , determine the value of $x$ .

Show answer & explanation

Answer: 5

Answer

The value of

x

is 5.

Rewriting 27 as

3^3

and 9 as

3^2

transforms the given equation into

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

. Equating exponents gives

3x - 3 = 2x + 2

, which simplifies directly to

x = 5

.

Step-by-Step Solution

1

Express numbers in terms of a common base

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

Both 27 and 9 are powers of 3, allowing reduction to a single base.

2

Apply power of a power index law

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

Multiply the base power by the expression in the exponent:

3 \times (x - 1) = 3x - 3

and

2 \times (x + 1) = 2x + 2

.

3

Equate exponents and solve for

x

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

Equal bases imply that the index powers must be equal.

Key Concept

Equating exponential expressions using a common base

Question 15Question →

If $2^{3x - 1} = 32$ , find the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2.

Rewriting 32 as

2^5

allows equating the exponents

3x - 1 = 5

. Solving for

x

gives

3x = 6

, which simplifies to

x = 2

.

Step-by-Step Solution

1

Express both sides of the equation using a common base of 2

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

The number 32 can be rewritten in index form as

2^5

.

2

Equate the indices

3x - 1 = 5

If

a^m = a^n

for a non-zero base

a \neq 1

, then

m = n

.

3

Solve the linear equation for

x

x = 2

Adding 1 to both sides yields

3x = 6

, and dividing by 3 gives

x = 2

.

Key Concept

Solving exponential equations by expressing numbers with equal bases

Question 16Question →

What is the product of all real values of

x

that satisfy the exponential equation

4^{x+1} - 17 \times 2^x + 4 = 0

?

-4

1

-1

4

Show answer & explanation

Answer: $-4$

Answer

The product of all real values of

x

satisfying the equation is

-4

.

Expressing

4^{x+1}

as

4(2^x)^2

allows setting

y = 2^x

, giving

4y^2 - 17y + 4 = 0

. Solving this quadratic equation yields

y = \frac{1}{4}

and

y = 4

. Solving

2^x = \frac{1}{4} = 2^{-2}

gives

x = -2

, and

2^x = 4 = 2^2

gives

x = 2

. The product of these two real solutions is

(-2) \times 2 = -4

.

Step-by-Step Solution

1

Rewrite

4^{x+1}

using index laws

4^{x+1} = 4^x \times 4^1 = (2^2)^x \times 4 = 4 \times (2^x)^2

Converting all exponential terms to base

2

allows substitution into a quadratic form.

2

Substitute

y = 2^x

into the equation

4y^2 - 17y + 4 = 0

This transforms the exponential equation into a standard quadratic equation in terms of

y

.

3

Solve the quadratic equation for

y

(4y - 1)(y - 4) = 0 \implies y = \frac{1}{4} \text{ or } y = 4

Factoring

4y^2 - 16y - y + 4 = 0

yields the two valid values for

y

.

4

Solve for

x

using

y = 2^x

For

y = \frac{1}{4}

:

2^x = 2^{-2} \implies x_1 = -2

. For

y = 4

:

2^x = 2^2 \implies x_2 = 2

.

Applying the law of indices

a^m = a^n \implies m = n

determines the real solutions for

x

.

5

Calculate the product of the solutions

x_1 \times x_2 = (-2) \times 2 = -4

The question asks for the product of the real values of

x

.

Key Concept

Quadratic form exponential equations and index transformation rules

Question 17Question →

What is the simplified numerical value of $27^{\frac{2}{3}} \times 9^{-\frac{1}{2}}$ ?

Show answer & explanation

Answer: 3

Answer

The simplified numerical value of the expression is 3.

Evaluating

27^{\frac{2}{3}}

gives

(\sqrt[3]{27})^2 = 3^2 = 9

. Evaluating

9^{-\frac{1}{2}}

gives

\frac{1}{\sqrt{9}} = \frac{1}{3}

. Multiplying

9

by

\frac{1}{3}

yields

3

.

Step-by-Step Solution

1

Rewrite bases in terms of prime factors

27 = 3^3

and

9 = 3^2

Expressing bases in power-of-3 form allows direct application of index laws.

2

Simplify each indexed term using

(a^m)^n = a^{m \cdot n}

and

a^{-n} = \frac{1}{a^n}

27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^2 = 9

and

9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1} = \frac{1}{3}

Fractional indices denote roots and negative indices denote reciprocals.

3

Multiply the resulting numbers

9 \times \frac{1}{3} = 3

Simplifying the final product yields the single numeric answer.

Key Concept

Fractional and Negative Index Laws

Question 18Question →

If $2^{x-1} + 2^{x+1} = 160$ , what is the value of $3^{x-2}$ ?

27

81

243

12

Show answer & explanation

Answer: 81

Answer

81

Factoring

2^{x-1}

from the left-hand side of

2^{x-1} + 2^{x+1} = 160

yields

2^{x-1}(1 + 2^2) = 160

, which reduces to

5 \cdot 2^{x-1} = 160

. Dividing by

5

gives

2^{x-1} = 32 = 2^5

. Equating exponents gives

x - 1 = 5

, so

x = 6

. Substituting

x = 6

into

3^{x-2}

gives

3^{6-2} = 3^4 = 81

.

Step-by-Step Solution

1

Factor out the common term

2^{x-1}

from the expression

2^{x-1} + 2^{x+1}

.

2^{x-1}(1 + 2^2) = 160

Applying index law

2^{x+1} = 2^{x-1} \cdot 2^2

allows factoring out

2^{x-1}

.

2

Simplify the bracketed terms and solve for

2^{x-1}

.

5 \cdot 2^{x-1} = 160 \implies 2^{x-1} = 32

Dividing both sides by

5

isolates the term with the unknown exponent.

3

Express

32

as a power of

2

and solve for

x

.

2^{x-1} = 2^5 \implies x - 1 = 5 \implies x = 6

Equating exponents since the bases are identical.

4

Substitute

x = 6

into the target expression

3^{x-2}

.

3^{6-2} = 3^4 = 81

Evaluating the power gives the final answer.

Key Concept

Factoring exponential expressions with common bases and applying laws of indices

Estimated Time:1m 30s

Question 19Question →

If $2^{2x + 1} - 5(2^x) + 2 = 0$ , what are the possible values of $x$ ?

-1

or

1

\frac{1}{2}

or

2

-1

or

2

-2

or

1

Show answer & explanation

Answer: $-1$ or $1$

Answer

-1

or

1

Using the addition law of indices,

2^{2x+1}

can be rewritten as

2^1 \cdot 2^{2x} = 2 \cdot (2^x)^2

. Substituting

y = 2^x

produces the quadratic equation

2y^2 - 5y + 2 = 0

, which factorizes into

(2y - 1)(y - 2) = 0

. This gives

y = \frac{1}{2}

or

y = 2

. Converting back to exponential equations gives

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

and

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

. Thus, the solutions for

x

are

-1

or

1

.

Step-by-Step Solution

1

Apply index laws to express the equation in terms of

2^x

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

, so the equation becomes

2 \cdot (2^x)^2 - 5(2^x) + 2 = 0

Splitting the exponent using

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

reveals a quadratic structure in

2^x

.

2

Substitute

y = 2^x

and solve the resulting quadratic equation

2y^2 - 5y + 2 = 0 \implies (2y - 1)(y - 2) = 0 \implies y = \frac{1}{2} \text{ or } y = 2

Factoring the quadratic expression yields the values for the substitution variable

y

.

3

Equate

2^x

to each solution of

y

to solve for

x

For

y = \frac{1}{2}

:

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

. For

y = 2

:

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

.

Using the negative index rule

a^{-n} = \frac{1}{a^n}

allows matching exponents when bases are identical.

Key Concept

Quadratic Equations Reducible to Index Form

Question 20Question →

If

x > 0

satisfies the exponential equation

3^{x+1} + 3^{1-x} = 10

find the value of

8^x + 4^{x-1}

.

Show answer & explanation

Answer: 9

Answer

The value of

8^x + 4^{x-1}

is 9.

Applying index laws to

3^{x+1} + 3^{1-x} = 10

yields

3(3^x) + \frac{3}{3^x} = 10

. Substituting

u = 3^x

gives

3u^2 - 10u + 3 = 0

, which factors to

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

, yielding

u = 3

or

u = \frac{1}{3}

. Thus

x = 1

or

x = -1

. Given

x > 0

,

x = 1

. Substituting

x = 1

into

8^x + 4^{x-1}

gives

8^1 + 4^0 = 8 + 1 = 9

.

Step-by-Step Solution

1

Apply the product and negative power laws of indices to separate the terms in the given equation.

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

and

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

, making the equation

3(3^x) + \frac{3}{3^x} = 10

.

According to the laws of indices,

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

and

a^{-n} = \frac{1}{a^n}

.

2

Substitute

u = 3^x

into the equation and clear the fraction to form a standard quadratic equation.

3u + \frac{3}{u} = 10 \implies 3u^2 - 10u + 3 = 0

.

Multiplying through by

u

eliminates the fraction and forms a quadratic in terms of

u

.

3

Factor the quadratic equation

3u^2 - 10u + 3 = 0

to find the values of

u

.

(3u - 1)(u - 3) = 0 \implies u = 3

or

u = \frac{1}{3}

.

Factoring by splitting the middle term gives the linear factors.

4

Equate

3^x

to the values of

u

and apply the constraint

x > 0

.

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

and

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

. Selecting the positive root gives

x = 1

.

Equating exponents with matching base 3 gives the solutions for

x

.

5

Substitute

x = 1

into the target expression

8^x + 4^{x-1}

and simplify.

8^1 + 4^{1-1} = 8 + 4^0 = 8 + 1 = 9

.

By the zero index law, any non-zero base raised to the power 0 equals 1 (

a^0 = 1

).

Key Concept

Solving quadratic-form exponential equations using index laws and applying the zero index rule.

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