Exponents

We prime factorize the number

1728 = 3³× 2⁶

⇒ 1728 = 3³× 4³

Since the powers are same so the bases can be multiplied

⇒ 1728 = 12³

We prime factorize the number 512

512 = 2⁹

Since reciprocal of a number results in negative power,

So = 2^-9

We prime factorize the number 169

169 = 13²

So we can say that

0.000169 = 13²× 10^-6

⇒ 0.000169 = (13× 10^-3)²

⇒ 0.000169 = 0.013²

The number can be expanded by expressing the number in terms of its digits and multiplying each digit with 10 raised to a specific power according to its place value

12345 = 1×10⁴ + 2×10³ + 3×10² + 4×10 + 5

Since the power is negative so

Since the bases are same so the powers can be added

3¹ × 3² × 3³ × 3⁴ × 3⁵ × 3⁶ = 3^{(1 + 2 + 3 + 4 + 5 + 6)}

⇒ 3¹ × 3² × 3³ × 3⁴ × 3⁵ × 3⁶ = 3²¹

Since there are two bases so their powers can be separately added.

2² × 3³ × 2⁴ × 3⁵ × 3⁶ = 2^{(2 + 4)} × 3^{(3 + 5 + 6)}

⇒ 2² × 3³ × 2⁴ × 3⁵ × 3⁶ = 2⁶ × 3¹⁴

Since the bases are same so the powers can be added

10⁴ × 10³ × 10² × 10 = 10^{(4 + 3 + 2 + 1)}

⇒ 10⁴ × 10³ × 10² × 10 = 10¹⁰

Since the power of 10 is 10 so there are 10 zeroes.

5³ × 5⁴ × 5⁵ × 5⁶ = 5^{(3 + 4 + 5 + 6)}

⇒ 5³ × 5⁴ × 5⁵ × 5⁶ = 5¹⁸

5⁷ × 5⁸ = 5^{(7 + 8)}

⇒ 5⁷ × 5⁸ = 5¹⁵

Since the power of 5 in 5³ × 5⁴ × 5⁵ × 5⁶ is greater so 5³ × 5⁴ × 5⁵ × 5⁶ is greater than 5⁷ × 5⁸

Since the bases are same so their powers are solved algebraically

10⁻¹ × 10² × 10⁻³ × 10⁴ × 10⁻⁵ × 10⁶ = 10^{(-1 + 2-3 + 4-5 + 6)}

On solving we get,

⇒ 10⁻¹ × 10² × 10⁻³ × 10⁴ × 10⁻⁵ × 10⁶ = 10³

Since the denominator and numerator have common base so their powers can be subtracted

We divide the two expressions

Since the denominator and numerator have common base so their powers can be subtracted

Since the fraction is greater than 1 so (3⁴ × 2³) is greater than (2⁵ × 3²)

We divide the two expressions

Since the denominator and numerator have common base so their powers can be subtracted

Since m and n are two distinct integers so the power can never become 0 and hence the fraction can never an integer.

Since is an integer so

the numerator must be greater than denominator

So b has to be either 1,2 or 4 to become an integer.

2^{m + n} = 2^mn

⇒ m + n = mn

Let m = n

So we get

2m = m²

⇒ m = 2

Since m = n, so

n = 2

(2^m)⁴ = 4⁶

⇒ (4^m)² = 4⁶

Since the bases are same so their powers can be equated

2m = 6

⇒ m = 3

Since the bases are common in numerator and denominator so their powers are subtracted

Since the bases are common in numerator and denominator so their powers are subtracted

Since after reducing the expression there is a 5 in the denominator but no multiple of 5 in the numerator so it is not an integer

We divide the two numbers

Since the numerator is greater than the denominator so 100⁴ is greater than 125³

Since bases are same so the powers can be subtracted

(1.8)⁶ × (4.2)^-3 = (18×10^-1)⁶×(42×10^-1)^-3

⇒ (1.8)⁶ × (4.2)^-3 = (3²×2×10^-1)⁶×(7×3×2×10^-1)^-3

⇒ (1.8)⁶ × (4.2)^-3 = 3¹²×2⁶×10^-6×7^-3×3^-3×2^-3×10³

⇒ (1.8)⁶ × (4.2)^-3 = 3⁹×2³×7^-3×10^-3

Since the bases are same so the powers can be equated

2m = m²

⇒ m = 2

So it can happen when m = 2

(3^m)ⁿ = 3^m × 3ⁿ

⇒ 3^mn = 3^{m + n}

Since the bases are same so the powers can be equated

mn = m + n

Let m = n

⇒ m² = 2m

⇒ m = 2

Answer: m = 2, n = 2

12⁶ = (4×3)⁶

⇒ 12⁶ = 2¹²×3⁶

On prime factorizing 162 we get

162 = 2× 3⁴

(3²)² = 9² = 81

((−2)³)² = (-8)² = 64

( − (5²))² = (-25)² = 625

So the value of the expression is

(3²)² − ((−2)³)² − ( − (5²))² = 81-64-625 = -608

The value of any number raised to the power zero is always 1

((0.6)²)⁰ = 1

(4.5)⁰ = 1

⇒ 1^-2 = 1

So the solution of the expression is

((0.6)²)⁰ − ((4.5)⁰)⁻² = 1-1 = 0

(4^-1)⁴ = 4^-4

⇒ (4^-1)⁴ = 2^-8

(8^-2)⁵ = (2^-6)⁵

⇒ (8^-2)⁵ = 2^-30

(64²)³ = (2¹²)³

(64²)³ = 2³⁶

= 2^-8×2⁵×2^-12×2^-30×2³⁶

⇒ = 2^-9

Using the third law of exponents, If a ≠0 is a number and m, n are integers, then(a^m)ⁿ= a^mn

So, (3^m)ⁿ = 3^mn

For a number a ≠ 0, we define

It is given that x,y, are nonzero real numbers,

Given 2^x − 2^x−2 = 192

⇒2^x–2^x.2⁻² = 192

Taking 2^x common from both the terms,

⇒2^x (1 - 2⁻²) = 192

For a number a ≠ 0and natural number n, we define

⇒2^x = 64× 4

⇒2^x = 2⁶ × 2²

For any a ≠ 0, and integers m,n, a^m × aⁿ = a^m+n

⇒2^x = 2⁸

Hence, x = 8

Using the third law of exponents, If a ≠0 is a number and m, n are integers, then (a^m)ⁿ= a^mn

For a number a ≠ 0, we define

For any a ≠ 0, and integers m,n, a^m × aⁿ = a^m+n

Given mⁿ =25

⇒mⁿ = 5²

⇒ (m, n) = (5, 2)

Hence, only 1 such pair of positive integers is possible.

Given the diameter of the Sun = 1.4 × 10⁹ meter

And the diameter of the Earth is = 1.2768 × 10⁷ meters

Ratio of the diameter of the sun and that of the earth =

For any number a ≠ 0 and positive integers m and n, not necessarily distinct,

Ratio of the diameter of the sun and that of the earth =

Ratio of the diameter of the sun and that of the earth =

{Using For a number a ≠ 0, we define

}

Using the third law of exponents, If a ≠0 is a number and m, n are integers, then (a^m)ⁿ= a^mn

Using For a ≠0 and b ≠ 0, and integer m,

(a × b)^m = a^m × b^m

Using For any a ≠ 0, and integers m,n,

a^m × aⁿ = a^m+n

Using For a ≠0 and b ≠ 0, and integer m,

(a × b)^m = a^m × b^m

Using For any a ≠ 0, and integers m, n,

a^m × aⁿ = a^m+n

2³ × 5⁴ × 20⁵ = 2³ × 5⁴ × (4× 5)⁵

Using For a ≠0 and b ≠ 0, and integer m,

(a × b)^m = a^m × b^m

⇒2³ × 5⁴ × 20⁵ = 2³ × 5⁴ × 4⁵× 5⁵

⇒2³ × 5⁴ × 20⁵ = 2³ × 5⁴ × (2² )⁵× 5⁵

Using third law of exponents, If a ≠0 is a number and m, n are integers, then (a^m)ⁿ= a^mn

⇒2³ × 5⁴ × 20⁵ = 2³ × 5⁴ × 2¹⁰× 5⁵

Using For any a ≠ 0, and integers m, n,

a^m × aⁿ = a^m+n

⇒2³ × 5⁴ × 20⁵ = 2³⁺¹⁰ × 5⁴⁺⁵

⇒2³ × 5⁴ × 20⁵ = 2¹³ × 5⁹

We know that 2× 5 = 10

So, 2³ × 5⁴ × 20⁵ = (2⁹ × 5⁹)× 2⁴

⇒2³ × 5⁴ × 20⁵ = 10⁹ × 16

⇒2³ × 5⁴ × 20⁵ =16000000000

Hence, there are 11 digits in the given number.

Given: a⁷ = 3

{Using the third law of exponents, If a ≠0 is a number and m, n are integers, then (a^m)ⁿ= a^mn}

{UsingFor any a ≠ 0, and integers m, n,

a^m × aⁿ = a^m+n}

For any number a ≠ 0 and positive integers m and n, not necessarily distinct,

Given: 2^m × a² = 2⁸

⇒2^m × a² = 2⁸× 1²

Then a + m = 8 + 1 = 9

Also, 2^m × a² = 2⁴ × 4²

Then a + m = 4 +4 = 8

Similarly, 2^m × a² = 2²× 8²

Then a + m = 2 +8 = 10

Given: 3^k × b² = 6⁴

⇒3^k × b² = (2× 3)⁴

Using For a ≠0 and b ≠ 0, and integer m,

(a × b)^m = a^m × b^m

⇒3^k × b² = 2⁴× 3⁴

⇒3^k × b² = 4²× 3⁴

On comparing,

k = 4, b = 4

⇒ k+b = 8

Using the third law of exponents, If a ≠0 is a number and m, n are integers, then (a^m)ⁿ= a^mn

Using For any a ≠ 0, and integers m, n,

a^m × aⁿ = a^m+n

Let the money he have be Rs 5^x and that he gave to his friend be Rs 5^y, such that x>y.

According to the question,

5^x - 5^y = 500

We can see from the equation that 5^x> 500 because 500 is the money that he is left with.

So 5^x = 625

⇒ x = 4

Then 625 - 5^y = 500

⇒5^y = 125

⇒ y = 3

So, the money he had = Rs 625 and that he gave to his friend = Rs 125