Surds, Indices and Logarithm

Question: If 2^x = 3^y = 6^-z, find the value of (1/x) + (1/y) + (1/z).

Solution: 
Let, 
2^x = 3^y = 6^-z = k
Now, 
2^x = k 
2 = k^1/x ......(1)
Similarly, 
3 = k^1/y ....(2)
And
6 = k^-1/z
⇒ 2 × 3 = k^-1/z
⇒ k^1/x × k^1/y = k^-1/z  ; [From (1) and (2)]
⇒ k^{(1/x + 1/y)} = k^-1/z
⇒ (1/x + 1/y) = - 1/z
∴ 1/x + 1/y + 1/z = 0

Question: If log_x (1/√32) = - 5/2 what is the value of x?

সমাধান:
দেওয়া আছে,
log_x (1/√32) = - 5/2
⇒ x^{- 5/2} = 1/√32 [log_aM = x হলে, a^x = M হয়]
⇒ x^{- 5/2} = 1/(32^1/2)
⇒ x^{- 5/2} = 32^{- 1/2}
⇒ x^{- 5/2} = (2⁵)^{- 1/2}
⇒ x^{- 5/2} = 2^{- 5/2}
∴ x = 2

(7^a)(7^b) = 7^c/7^d
Or, 7^{a + b} = 7^c-d
Or, a + b = c - d
Or, d = c - a - b

Question: If log_m(81) = 4, what is the value of m?

Solution:
log_m(81) = 4
⇒ m⁴ = 81 [log_b(a) = c ⇒ b^c = a]
⇒ m⁴ = 3⁴
∴ m = 3

Question: 2³⁰ + 2³⁰ + 2³⁰ + 2³⁰ = ?

Solution
2³⁰ + 2³⁰ + 2³⁰ + 2³⁰
= 4 × 2³⁰
= 2² × 2³⁰
= 2^{2 + 30}
= 2³²

log₁₀4+log₁₀25
= log₁₀(4×25)
= log₁₀100
= log₁₀10²
= 2log₁₀10
= 2

Question: If x³ = 125, then x² + x = ?

Solution: 
Given that, 
x³ = 125
⇒ x³ = 5³
∴ x = 5

Now, 
x² + x = 5² + 5 = 25 + 5 = 30

So the value of x² + x is 30

Question:

Solution:

 Here, e^x = 5
⇒ lne^x = ln 5
⇒ xlne = ln 5 [As, lne = 1]
∴ x = 1.61 

Question: 


Solution: 

Question: If x and y are positive integers such that 5^{x + y} = 625 and 4^{x - y} = 16, what is the value of x² - y²?

Solution:
Given that,
5^{x + y} = 625
⇒ 5^{x + y} = 5⁴
∴ x + y = 4 ........(1)

And,
4^{x - y} = 16
⇒ 4^{x - y} = 4²
∴ x - y = 2 ....... (2)

Now (1) + (2), we get,
⇒ (x + y) + (x - y) = 4 + 2
⇒ 2x = 6
⇒ x = 6 / 2
∴ x = 3

From (1),
3 + y = 4
⇒ y = 4 - 3
∴ y = 1

∴ x² - y² = (3)² - (1)²
= 9 - 1
= 8

Question: log₃(x² + 3x) - log₃(x + 3) = 2, Then what is the value of x?

Solution:
Given that, 
log₃(x² + 3x) - log₃(x + 3) = 2
⇒ log₃[(x² + 3x)/(x + 3)] = 2 ; [log₃A - log₃B = log₃(A/B)] 
⇒ log₃[x(x + 3)/(x + 3)] = 2 
⇒ log₃x = 2
⇒ x = 3² 
∴ x = 9

Question: Solve for x;  log_x3 + log_x9 + log_x27 + log_x81 = 10.

Solution: 
Given that, 
log_x3 + log_x9 + log_x27 + log_x81 = 10
⇒ log_x(3 × 9 × 27 × 81) = 10
⇒ log_x(3¹ × 3² × 3³ × 3⁴) = 10
⇒ log_x(3¹⁰) = 10
⇒ 10 log_x3 = 10
⇒ log_x3 = 10/10
⇒ log_x3 = 1
⇒ x¹ = 3
∴ x = 3

Question: Find the value of x if log_x 324 = 4.

Solution:
log_x324 = 4
⇒ x⁴ = 324
⇒ (x²)² = 18²
⇒ x² = 18²
⇒ x = √18
⇒ x = √3² × 2
⇒ x = 3√2

By definition x ≠ 1, =1
Given equation can be written as
log(x+1)² / x²−1=log2
=>x+1 / x−1=2
=>x=3

Question: If 5^{x + y} = 25 and 5^{x - y} = 5, then what are the values of x and y respectively?

Solution:
Given,
5^{x + y} = 25
⇒ 5^{x + y} = 5²
⇒ x + y = 2 .......(1)

Again,
5^{x - y} = 5
⇒ 5^{x - y} = 5¹
⇒ x - y = 1 ........(2)

Now, solving (1) and (2) we get,
x + y + x - y = 2 + 1
⇒ 2x = 3
∴ x = 3/2

Now,
x + y = 2
⇒ 3/2 + y = 2
⇒ y = 2 - (3/2)
​⇒ y = 1/2
​
​(x, y) = (3/2, 1/2)

log_a10−log_a(10/a)
=log_a10−[log_a10−log_aa]
=log_a10−log_a10+log_aa
=0+1=1

Question: Find the value of n, if 27^{{n - (1/3)}} = 243.

Solution:
27^{{n - (1/3)}} = 243
⇒ (3³)^{{n - (1/3)}} = 3⁵
⇒ 3^{(3n - 1)} = 3⁵
⇒ 3n - 1 = 5
⇒ 3n = 5 + 1
⇒ 3n = 6
∴ n = 2

Question: If log⁡_m243 + log⁡_m81 = 9, find the value of m.
​
​Solution:
​Given that,
​log⁡_m243 + log⁡_m81 = 9
​⇒ ​​log⁡_m(243 × 81) = 9
⇒ ​​log⁡_m19683 = 9
⇒ ​m⁹ = 19683
​⇒ ​m⁹ = 3⁹
∴ m = 3​

Question: Given 16^x = 4⁴, what is x?

Solution:
16^x = 4⁴
⇒ (4²)^x = 4⁴
⇒ 4^2x = 4⁴
⇒ 2x = 4
⇒ x = 4/2
∴ x = 2

Question: If log_x(125/8) = - 3, what is the value of x?

Solution:
log_x(125/8) = - 3
⇒ x^{- 3} = 125/8 [log_b(a) = c ⇒ b^c =a]
⇒ x^{- 3} = 5³/2³
⇒ x^{- 3} = (5/2)³
⇒ x^{- 3} = (2/5)^{- 3}
∴ x = 2/5

Question: If x and y are positive real numbers, then (2x⁰ - 5y⁰)⁴ = ?

Solution:
We know that for any positive real number,
x⁰ = 1 and y⁰ = 1

So, (2x⁰ - 5y⁰)⁴
= (2 × 1 - 5 × 1)⁴
= (2 - 5)⁴
= (- 3)⁴
= 81

The value of (2x⁰ - 5y⁰)⁴ is 81

2¹²ⁿ−6⁴ⁿ
=(2³)⁴ⁿ−6⁴ⁿ
=8⁴ⁿ−6⁴ⁿ
=(8²)²ⁿ−(6²)²ⁿ
= 64²ⁿ−36²ⁿ (n=1)
= 64²−36²
=(64+36)(64−36)
=100×28 = 2800