CAT — Logarithms

Given, log_x(x² + 12) = 4

Using the definition of logarithm, x⁴ = x² + 12

Rearrange, x⁴ − x² − 12 = 0

Let t = x²

Then, t² − t − 12 = 0

Factorizing, (t − 4)(t + 3) = 0

Since x is positive, t = 4

Therefore, x² = 4 → x = 2

Now, 3 log_y x = 1 → log_y x = 1/3

Using the definition of logarithm, y^(1/3) = x

Substitute x = 2, y^(1/3) = 2

Cubing both sides, y = 8

Therefore, x + y = 2 + 8 = 10

Answer: C. 10

Given:
a, b and c are positive real numbers such that
a > 10 ≥ b ≥ c
and
log₈(a + b)/log₂c + log₂₇(a − b)/log₃c = 2/3
Find the greatest possible integer value of a.

Step 1: Simplify the logarithmic expression
Using the change of base formula,
log₈(a + b)
= log₂(a + b)/3
and
log₂₇(a − b)
= log₃(a − b)/3
Hence,
(1/3) logc(a + b) + (1/3) logc(a − b) = 2/3
Multiplying by 3,
logc(a + b) + logc(a − b) = 2
Using the product rule of logarithms,
logc[(a + b)(a − b)] = 2
Therefore,
a² − b² = c²

Step 2: Use the given constraints
Since
b ≤ 10
and
c ≤ 10,
we have
b² ≤ 100
and
c² ≤ 100
Thus,
a² = b² + c² ≤ 200
Hence,
a ≤ √200 ≈ 14.14
Therefore, the greatest possible integer value of a can be at most 14.

Step 3: Check whether a = 14 is possible
Let
a = 14
and
b = 10.
Then,
c² = a² − b²
= 196 − 100
= 96
So,
c = √96 = 4√6 ≈ 9.80
This satisfies
10 ≥ b ≥ c
since
10 ≥ 9.80.
Hence, all the given conditions are satisfied.

Therefore, the greatest possible integer value of a is
14

Answer: 14

Step 1: Simplify the second term logₓ(25) / logₓ(0.008) = log(25) / log(0.008) 25 = 5², 0.008 = 5⁻³ = (2 log 5) / (−3 log 5) = −2/3 Step 2: Simplify the equation log√3(x) − 2/3 = 16/3 log√3(x) = 18/3 = 6 Step 3: Convert to base 3 log√3(x) = log₃(x) / log₃(√3) = log₃(x) / (1/2) = 2 log₃(x) 2 log₃(x) = 6 → log₃(x) = 3 → x = 27 Step 4: Find the required value log₃(3x²) = log₃(3 × 27²) = log₃(3 × 3⁶) = log₃(3⁷) = 7

Since the base 1/4 < 1, log₁/₄(x) > 0 implies 0 < x < 1. So we need 0 < n² − 7n + 11 < 1. From n² − 7n + 11 < 1: n² − 7n + 10 < 0 → (n−2)(n−5) < 0 → 2 < n < 5. Integer candidates: n = 3, 4. Checking domain (n² − 7n + 11 > 0): for n = 3: 9 − 21 + 11 = −1 < 0. For n = 4: 16 − 28 + 11 = −1 < 0. Neither satisfies the domain condition. Answer: 0.

Step 1: Convert all logarithms to base 2. Since 64 = 2⁶, 8 = 2³, 512 = 2⁹: log₆₄ x² = (1/6) log₂ x² = (1/3) log₂ x. log₈ √y = (1/3) log₂ √y = (1/6) log₂ y. 3 log₅₁₂ (√yz) = 3 × (1/9) log₂ (√yz) = (1/3) × (1/2) log₂ (yz) = (1/6)(log₂ y + log₂ z). Substituting: (1/3) log₂ x + (1/6) log₂ y + (1/6) log₂ y + (1/6) log₂ z = 4. So, (1/3) log₂ x + (1/3) log₂ y + (1/6) log₂ z = 4. Multiply throughout by 6: 2 log₂ x + 2 log₂ y + log₂ z = 24. Using logarithm properties: log₂ (x²y²z) = 24. Hence, x²y²z = 2²⁴.

Step 2: Apply AM-GM Inequality. Apply AM-GM directly to x², y² and z: (x² + y² + z)/3 ≥ (x²y²z)^(1/3) = 2⁸ = 256. Equality holds when x² = y² = z = 256. Since x and y are positive: x = 16, y = 16, z = 16. Therefore, x + y + z = 16 + 16 + 16 = 48.

Given:
10ˣ + 4/10ˣ = 81/2
Find the sum of all distinct real values of x.

Step 1: Substitute a variable
Let
10ˣ = t
where
t > 0.
Then,
t + 4/t = 81/2

Step 2: Form a quadratic equation
Multiply both sides by 2t,
2t² + 8 = 81t
2t² − 81t + 8 = 0

Step 3: Solve the quadratic
Factorizing,
(2t − 1)(t − 8) = 0
Hence,
t = 1/2
or
t = 8
Since
t = 10ˣ,
the corresponding values of x are
x = log₁₀(1/2)
and
x = log₁₀8

Step 4: Find the sum of the values
Sum
= log₁₀(1/2) + log₁₀8
= log₁₀(8/2)
= log₁₀4
= log₁₀(2²)
= 2 log₁₀2

Answer:
A. 2 log₁₀2

Given:
3ᵃ = 4
4ᵇ = 5
5ᶜ = 6
6ᵈ = 7
7ᵉ = 8
8ᶠ = 9
Find the value of
abcdef.

Step 1: Express each variable using logarithms
From the given equations,
a = log₃4
b = log₄5
c = log₅6
d = log₆7
e = log₇8
f = log₈9

Step 2: Write the product
abcdef
= log₃4 × log₄5 × log₅6 × log₆7 × log₇8 × log₈9

Step 3: Use the property of logarithms
Using
logₘn × logₙp = logₘp,
we get
log₃4 × log₄5 = log₃5
Then,
log₃5 × log₅6 = log₃6
Continuing similarly,
abcdef
= log₃9

Step 4: Evaluate the final logarithm
Since
3² = 9,
log₃9 = 2

Final Answer
2

Step 1: Convert all logarithms to base 10
log₁₀₀x = log₁₀x / 2
log₁₀₀₀x = log₁₀x / 3
Let log₁₀x = y.
4y + 4(y/2) + 8(y/3) = 13

Step 2: Solve for y
4y + 2y + 8y/3 = 13
Multiply by 3: 12y + 6y + 8y = 39 → 26y = 39 → y = 3/2

Step 3: Find x
x = 10^(3/2) = 10 × √10 ≈ 31.62

Step 4: Find the greatest integer not exceeding x
⌊x⌋ = 31

Step 1: Domain conditions
x − 3 > 0 and x − 3 ≠ 1 → x > 3 and x ≠ 4.

Step 2: Rewrite using 2 = log₍ₓ₋₃₎(x−3)²
log₍ₓ₋₃₎(x² − 9) = log₍ₓ₋₃₎[(x+1)(x−3)²]
→ x² − 9 = (x+1)(x−3)²

Step 3: Factor and simplify
(x−3)(x+3) = (x+1)(x−3)²
Cancel (x−3) [valid since x > 3]: x + 3 = (x+1)(x−3)

Step 4: Expand and solve
x + 3 = x² − 2x − 3 → x² − 3x − 6 = 0
x = (3 ± √33)/2

Step 5: Check domain
(3 − √33)/2 < 3 → rejected.
Valid solution: x = (3 + √33)/2