Surds, Indices and Logarithm

Question: If 4^{n - 2} = 128, find the value of n.

Solution:
4^{n - 2} = 128
⇒ (2²)^{n - 2} = 2⁷
⇒ 2^{2(n - 2)} = 2⁷
⇒ 2^{2n - 4} = 2⁷
⇒ 2n - 4 = 7
⇒ 2n = 7 + 4
⇒ 2n = 11
⇒ n = 11/2
∴ n = 5.5

Question: If log_x(1/81^-1) = - 4, what is the value of x?

Solution:
দেওয়া আছে,
log_x(1/81^-1) = - 4
⇒ log_x{1/(1/81)} = - 4 
⇒ log_x(81) = - 4 
⇒ x^{- 4} = 81 [log_ab = c ⇒ a^c = b]
⇒ x^{- 4} = 3⁴
⇒ (1/x)⁴ = 3⁴
⇒ 1/x = 3
∴ x = 1/3

Question: If log_x16 = 4/3, what is the value of x?

Solution:
log_x16 = 4/3
⇒ x^4/3 = 16 [log_ba = c ⇒ b^c = a]
⇒ (x^4/3)^3/4 = 16^3/4
⇒ x = 16^3/4
⇒ x = (2⁴)^3/4
⇒ x = 2³
∴ x = 8

Question: 


Solution:
Given that,

Question:

Solution:

Question: If log₃[log₂(log₂x)] = 1, then x is equal to = ?

Solution:
দেওয়া আছে, log₃[log₂(log₂(x)] = 1
⇒ log₂(log₂(x) = 3¹
⇒ log₂(log₂(x)= 3 ; [log_ab = c ⇒ b = a^c]
⇒ log₂(x) = 2³
⇒ log₂(x) = 8
⇒ x = 2⁸
∴  x = 256

Given,
x = y^a
⇒ x = (z^b)^a
⇒ x = (x^c)^ab
⇒ (x)^abc = x¹
∴ abc = 1

Question: If p and q are positive integers such that 2^p × 4^q = 32, then 2p + q =?

Solution:
Given, 2^p × 4^q = 32
⇒ 2^p × 2^2q = 2⁵
⇒ 2^{p + 2q} = 2⁵
⇒ p + 2q = 5

Since p and q are positive integers, test values of q :

If q = 1 then p = 3 
If q = 2 then p = 1
 
Now compute 2p + q :
For (p, q)=(3, 1)
2p + q= 2(3) + 1 = 7

For (p, q)=(1, 2)
2p + q = 2(1) + 2 = 4

(√8–√4–√2)
=2√2–2−√2
=2√2–√2–2
=√2–2

Solution:
Given that,
⇒ log₂(x + 3) = 4
⇒ x + 3 = 2⁴
⇒ x + 3 = 16
⇒ x = 16 - 3
x = 13

Question: If log_x(8/125) = - 3, then x = ?

Solution:
log_x(8/125) = - 3
⇒ x^{- 3} = 8/125
⇒ x^{- 3} = (2/5)³
⇒ x^{- 3} = (5/2)^{- 3}
⇒ x = 5/2
∴ x = 5/2

Question: If (1/3)^2y = 1/81, then find (0.3)^y = ?

Solution:
দেওয়া আছে,
(1/3)^2y = 1/81
⇒ (1/3)^2y = (1/3)⁴
⇒ 2y = 4
⇒ y = 4/2
⇒ y = 2

এখন,
(0.3)^y 
​= (0.3)² 
​= 0.09

∴ নির্ণেয় মান হলো 0.09

Question: (64)^0.20/(4)^0.10 = ?

Solution: 
Given that, 
(64)^0.20/(4)^0.10
= (4³)^0.20/(4)^0.10
= (4)^0.60/(4)^0.10
= (4)^{0.60 - 0.10}
= (4)^0.50
= (4)^1/2
= √4
= 2

Question: 


Solution: 

Question: If log_x(1/125) = - 3, then x = ?

Solution:
Given that,
log_x(1/125) = - 3
⇒ x^{- 3} = 1/125  [log_a(b) = c  ⇒ a^c = b]
⇒ 1/x³ = 1/125
⇒ x³ = 125
∴ x = 5

³√32 = 2^x
⇒ 2^5/3 = 2^x
⇒ x = 5/3

Question: 


Solution: 

Question: If log₁₀x + log₁₀y = 3 and log₁₀x - log₁₀y = 1, then x and y are respectively.

Solution: 
Given that, 
log₁₀x + log₁₀y = 3 ......(1)
log₁₀x - log₁₀y = 1 .......(2)

Now, (1) + (2) then we get,
⇒ log₁₀x + log₁₀y + log₁₀x - log₁₀y = 3 + 1
⇒ 2log₁₀x = 4
⇒ log₁₀x = 4/2 = 2
⇒ x = 10²
∴ x = 100

From (1) we get,
⇒ log₁₀x + log₁₀y = 3
⇒ log₁₀100 + log₁₀y = 3
⇒ 2 log₁₀10 + log₁₀y = 3
⇒ log₁₀y = 3 - 2
⇒ log₁₀y = 1
⇒ y = 10¹
∴ y = 10

∴ x and y are respectively 100 and 10.

Question: If log_x(16/81) = - 4, then what is the value of x?

Solution:
log_x(16/81) = - 4
⇒ x^{- 4} = 16/81 [log_ba = c ⇒ b^c = a]
⇒ x^{- 4} = (2/3)⁴
⇒ x^{- 4} = 1/(3/2)⁴
⇒ x^{- 4} = (3/2)^{- 4}
⇒ x = 3/2

Question:
 

Solution:

16^3/4
= (2⁴)^3/4
= 2^4×3/4
= 2³
= 8

Question: If log_xy = 100 and log₂x = 10, then the value of y is- 

Solution:
log₂x = 10
⇒ x = 2¹⁰ 

log_xy = 100 
⇒ y = x¹⁰⁰ 
= (2¹⁰)¹⁰⁰
= 2¹⁰⁰⁰

Question: Solve for x: log₂(x + 3) = 4

Solution: 
Given that, 
log₂(x + 3) = 4
⇒ x + 3 = 2⁴
⇒ x + 3 = 16
⇒ x = 16 - 3
∴ x = 13

Question: If log_a2 = a and log_a5 = b, then log_a50 = ?

Solution:
50 = 2 × 5²

log_a50 = log_a(2 × 5²) = log_a2 + log_a5²
∴ log_a50 = log_a2 + 2 log_a5
∴ log_a50 = a + 2b [log_a2 = a and log_a5 = b]

Question: 


Solution:

Question: What is the quotient when (x^{- 1} - 1) is divided by (x - 1)?

Solution: 

Question: Solve for x: log₃ (x - 1) = 3

Solution:
Given that,
log₃ (x - 1) = 3
⇒ x - 1 = 3³
⇒ x - 1 = 27
⇒ x = 27 + 1
∴ x = 28

Question: , then what is the value of m?

Solution:

We have 4^m > 1 
Now, if m = -1, then 4^m = 4 ^-1 = 1/4 = 0.25 < 1, so incorrect.
if m = 1, then 4^m = 4¹ = 4 > 1, correct.

∴ m = 1

log₁₀1000 
= log₁₀10³
= 3log₁₀10
= 3×1=3

Question: 

Solution: