CAT — Surds & indices

Step 1: Rewrite the equation using the same base. Given: 9^(x² + 2x − 3) − 4(3^(x² + 2x − 2)) + 27 = 0. Since 9 = 3², 9^(x² + 2x − 3) = 3^[2(x² + 2x − 3)] = 3^(2x² + 4x − 6). Also, 3^(2x² + 4x − 6) = (3^(x² + 2x − 3))². Let y = 3^(x² + 2x − 3). Then, 3^(x² + 2x − 2) = 3 × 3^(x² + 2x − 3) = 3y. Substitute into the equation: y² − 4(3y) + 27 = 0 → y² − 12y + 27 = 0.

Step 2: Solve the quadratic equation. y² − 12y + 27 = 0 → (y − 3)(y − 9) = 0. Therefore, y = 3 or y = 9.

Step 3: Find the corresponding values of x. Since y = 3^(x² + 2x − 3). Case 1: y = 3. 3^(x² + 2x − 3) = 3 → x² + 2x − 3 = 1 → x² + 2x − 4 = 0. Using the quadratic formula: x = (−2 ± √20)/2 = −1 ± √5. Case 2: y = 9. 3^(x² + 2x − 3) = 9 = 3² → x² + 2x − 3 = 2 → x² + 2x − 5 = 0. Using the quadratic formula: x = (−2 ± √24)/2 = −1 ± √6.

Step 4: Find the product of all possible values of x. For x² + 2x − 4 = 0, product of roots = −4. For x² + 2x − 5 = 0, product of roots = −5. Hence, the product of all four values is (−4) × (−5) = 20.

Given:
(a + b√3)² = 52 + 30√3,
where a and b are natural numbers.
Find the value of
a + b.

Step 1: Expand the left-hand side
(a + b√3)²
= a² + 2ab√3 + 3b²

Step 2: Compare the rational and irrational parts
Comparing both sides,
a² + 3b² = 52
2ab = 30
ab = 15

Step 3: Find the values of a and b
Since
ab = 15,
the possible pairs are
(1, 15), (3, 5), (5, 3), (15, 1).
Check each pair in
a² + 3b² = 52.
For
a = 5, b = 3,
25 + 27 = 52 ✓
All other pairs do not satisfy the equation.
Hence,
a = 5
and
b = 3.

Step 4: Find the required value
a + b
= 5 + 3
= 8

Answer:
B. 8

Given, √(5x + 9) + √(5x − 9) = 3(2 + √2)
Square both sides.
LHS = (√(5x + 9) + √(5x − 9))² = (5x + 9) + (5x − 9) + 2√[(5x + 9)(5x − 9)] = 10x + 2√(25x² − 81)
RHS = [3(2 + √2)]² = 9(2 + √2)² = 9(4 + 4√2 + 2) = 54 + 36√2
Therefore, 10x + 2√(25x² − 81) = 54 + 36√2
Divide by 2, 5x + √(25x² − 81) = 27 + 18√2
Since 18√2 is the irrational part, √(25x² − 81) = 18√2 and 5x = 27
Therefore, x = 27/5
Now find, √(10x + 9) = √[10 × (27/5) + 9] = √(54 + 9) = √63 = 3√7
Answer:
C. 3√7

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: Rewrite the equation
Let A = 2^(2x²) and B = 2^(x+15).
Then: 2^(4x²) = A², 2^(2x²+x+16) = 2AB, 2^(2x+30) = B²
The equation becomes: A² − 2AB + B² = 0

Step 2: Factorize
(A − B)² = 0 → A = B

Step 3: Equate the exponents
2x² = x + 15 → 2x² − x − 15 = 0

Step 4: Solve the quadratic
2x² − 6x + 5x − 15 = 0
2x(x − 3) + 5(x − 3) = 0
(x − 3)(2x + 5) = 0
x = 3 or x = −5/2

Step 5: Find the required sum
Sum = 3 + (−5/2) = 1/2

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

Step 1: Express each number as a power of a prime. 625 = 5⁴ and 128 = 2⁷. Therefore, (625)⁶⁵ × (128)³⁶ = (5⁴)⁶⁵ × (2⁷)³⁶ = 5²⁶⁰ × 2²⁵². Step 2: Simplify the expression. Write 5²⁶⁰ as 5²⁵² × 5⁸. So, 5²⁶⁰ × 2²⁵² = (5²⁵² × 2²⁵²) × 5⁸ = 10²⁵² × 5⁸. Step 3: Calculate 5⁸. 5⁸ = 390625. Hence, 10²⁵² × 5⁸ = 390625 × 10²⁵². This is the number 390625 followed by 252 zeros. Step 4: Find the sum of the digits. Sum of the digits of 390625 = 3 + 9 + 0 + 6 + 2 + 5 = 25. The trailing zeros do not affect the digit sum.

Step 1: Express all bases as powers of 2
8 = 2³ → 1/8 = 2^(−3)
32768 = 2¹⁵ → 1/32768 = 2^(−15)
Substituting: (2^(−3))^k × (2^(−15))^(1/3) = 2^(−3) × (2^(−15))^(1/k)

Step 2: Simplify the exponents
2^(−3k) × 2^(−5) = 2^(−3) × 2^(−15/k)
2^(−3k − 5) = 2^(−3 − 15/k)

Step 3: Equate the exponents
−3k − 5 = −3 − 15/k
3k + 5 = 3 + 15/k
3k + 2 = 15/k

Step 4: Form and solve the quadratic equation
3k² + 2k = 15 → 3k² + 2k − 15 = 0
Sum of roots = −b/a = −2/3

Given:
Three equal circles of radius r touch each other externally.
Two more circles, X and Y, are drawn such that each touches all the three circles.
Radius of X is greater than that of Y.
Find the ratio of the radii of X and Y.

Step 1: Locate the centres
The centres of the three equal circles form an equilateral triangle of side
2r.
The centre of both circles X and Y is the common centre (circumcentre) of this equilateral triangle.
The distance from this centre to each vertex is the circumradius of the equilateral triangle.
Circumradius
= (2r)/√3
= 2r/√3

Step 2: Find the radius of the larger circle X
Since X encloses the three circles, the distance between its centre and the centre of each small circle is
R − r,
where R is the radius of X.
Hence,
R − r = 2r/√3
R = r + 2r/√3
= r(1 + 2/√3)

Step 3: Find the radius of the smaller circle Y
Since Y lies inside the gap, the distance between its centre and the centre of each small circle is
r + y,
where y is the radius of Y.
Hence,
r + y = 2r/√3
y = 2r/√3 − r
= r(2/√3 − 1)

Step 4: Find the ratio
R : y
= (1 + 2/√3) : (2/√3 − 1)
Multiply both terms by √3,
= (√3 + 2) : (2 − √3)
Rationalizing,
= (√3 + 2)² : (4 − 3)
= (3 + 4 + 4√3) : 1
= (7 + 4√3) : 1

Answer:
A. (7 + 4√3) : 1

Step 1: Compute 10²⁵ − 123.
= 99999999999999999999999877
(22 nines followed by 877)

Step 2: Add 10⁵⁰.
The full number has:
- One leading 1
- Then 24 zeros
- Then 1
- Then 21 nines followed by 99877

Step 3: Sum of digits
= 1 + 1 + (21 × 9) + 9 + 9 + 8 + 7 + 7
= 2 + 189 + 40
= 221

Step 1: Find x + 1/x
x² + 1/x² = (x + 1/x)² − 2 = 25 → (x + 1/x)² = 27 → x + 1/x = 3√3

Step 2: x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)
= (3√3)³ − 3(3√3) = 81√3 − 9√3 = 72√3

Step 3: x⁵ + 1/x⁵ = (x³ + 1/x³)(x² + 1/x²) − (x + 1/x)
= 72√3 × 25 − 3√3 = 1800√3 − 3√3 = 1797√3

Step 4: x⁷ + 1/x⁷ = (x⁵ + 1/x⁵)(x² + 1/x²) − (x³ + 1/x³)
= 1797√3 × 25 − 72√3 = 44925√3 − 72√3 = 44853√3

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: Express in prime factors.
12 = 2²×3, 4 = 2², 8 = 2³, 20 = 2²×5, 243 = 3⁵

Step 2: Compare powers of 2
LHS: 2^(24x) × 2^(48x+24) = 2^(72x+24)
RHS: 2^(12z) × 2^(24x) = 2^(12z+24x)
72x + 24 = 12z + 24x → 48x + 24 = 12z → z = 4x + 2

Step 3: Compare powers of 3
LHS: 3^(12x), RHS: 3^(15x−30)
12x = 15x − 30 → 3x = 30 → x = 10
z = 4(10) + 2 = 42

Step 4: Compare powers of 5
LHS: 5^(2y), RHS: 5^(12x)
2y = 12x → y = 6x = 60

Step 5: x + y + z = 10 + 60 + 42 = 112

Given:
√(x + 6√2) − √(x − 6√2) = 2√2
Find x.

Step 1: Square both sides
[√(x + 6√2) − √(x − 6√2)]² = (2√2)²
x + 6√2 + x − 6√2 − 2√[(x + 6√2)(x − 6√2)] = 8
2x − 2√(x² − 72) = 8
Divide by 2,
x − √(x² − 72) = 4

Step 2: Isolate the square root
√(x² − 72) = x − 4
Since the left side is non-negative,
x ≥ 4.

Step 3: Square again
x² − 72 = (x − 4)²
x² − 72 = x² − 8x + 16
8x = 88
x = 11

Step 4: Verify the solution
Substituting x = 11,
√(11 + 6√2) − √(11 − 6√2)
= √9 − √1
= 3 − 1
= 2
= 2√2/√2
Since
11 + 6√2 = (3 + √2)²
and
11 − 6√2 = (3 − √2)²,
we have
√(11 + 6√2) = 3 + √2
√(11 − 6√2) = 3 − √2
Therefore,
(3 + √2) − (3 − √2)
= 2√2,
which satisfies the given equation.
Hence,
x = 11

Answer: 11

Let √x = a, √y = b, √z = c, where a, b and c are positive and distinct.
The given condition becomes
1/(b + c) = (1/2) [1/(a + c) + 1/(a + b)]
Multiply both sides by 2, 2/(b + c) = 1/(a + c) + 1/(a + b)
Taking LCM on the RHS, 2/(b + c) = [(a + b) + (a + c)] / [(a + c)(a + b)] = (2a + b + c) / [(a + b)(a + c)]
Cross-multiply, 2(a + b)(a + c) = (b + c)(2a + b + c)
Expand both sides, 2(a² + ab + ac + bc) = 2ab + 2ac + b² + 2bc + c²
Cancel the common terms, 2a² = b² + c²
Since a² = x, b² = y, c² = z, we get 2x = y + z or y + z = 2x
Hence, y, x and z are in arithmetic progression.
Answer:
B. y, x and z are in arithmetic progression

Step 1: Square both sides
(a + b√n)² = 29 − 12√5
a² + b²n + 2ab√n = 29 − 12√5

Step 2: Compare rational and irrational parts
a² + b²n = 29
ab√n = −6√5

Step 3: Eliminate √n
Squaring: a²b²n = 180
Using b²n = 29 − a²:
a²(29 − a²) = 180 → a⁴ − 29a² + 180 = 0
(a² − 20)(a² − 9) = 0
Since a is an integer: a² = 9 → a = ±3

Step 4: Find b²n
b²n = 29 − 9 = 20
Possible values: b² = 1, n = 20 or b² = 4, n = 5

Step 5: Check each case
Case 1: b² = 1, n = 20 → ab = −3
a = −3, b = 1: −3 + √20 > 0 ✓ → a + b + n = −3 + 1 + 20 = 18

Case 2: b² = 4, n = 5 → ab = −6
a = −3, b = 2: −3 + 2√5 > 0 ✓ → a + b + n = −3 + 2 + 5 = 4

Step 6: Maximum value = 18