Breakup G Strategy | DPAC Logarithm + Equations | Hard
The sum of all possible real values of x for which
logx-3(x² – 9) = logx-3(x + 1) + 2 is :
1.√33
2.-3
3.(3+√33)/2
4.3
Answer & Explanation
log(x-3) ((x²−9)/(x+1)) = 2
By definition of log :
(x²−9)/(x+1) = (x−3)²
Now equating :
(x²−9)/(x+1) = x²−6x+9
x²−9 = (x²−6x+9)(x+1)
Solve the cubic equation:
x³−6x²+3x+18 = 0
Factor the cubic equation:
(x−3)(x²−3x−6) = 0
Solve x²−3x−6 = 0 using the quadratic formula:
x = (3 ± √33)/2
Sum of solutions:
Sum = (3 + √33)/2
Option 3 is the answer
X Maro G Strategy | Surds and Indices | Moderate
If 12¹²x × 4²⁴x + ¹² × 5²y = 8⁴z × 20¹²x × 2433x-6 , where x, y and z are natural numbers, then x+y+z equals_____
Solution & Explanation
12¹²x × 4²⁴x + ¹² × 5²y = 8⁴z × 20¹²x × 2433x-6
12 = 2² × 3 , 4 = 2² , 8 = 2³ , 20 = 2² × 5 , 243 = 3⁵
Thus expression can be written as :
224x × 312x × 248x+24 × 52y = 212z × 2²⁴x × 512x × 315x – 30
Cancels out the same powers and comparing the powers
12x = 15x – 30 , x = 10
2y = 12x , 2y = 12(10) , y = 60
12 z = 48x + 24 , 12z = 504 , z = 42
x + y + z = 10 + 60 + 42 = 112
Answer 112
Breakup G Strategy | AP , GP + Equation | Moderate
In an arithmetic progression, if the sum of fourth, seventh and tenth term is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is : [TITA]
Solution & Explanation
Given : Sum of 4th, 7th, and 10th terms = 99
Sum of first 14 terms = 497
Use the formulas for AP
(a + 3d) + (a + 6d) + (a + 9d) = 99
This simplifies to:
a + 6d = 33 (Equation 1)
(14/2) × [2a + 13d] = 497
2a + 13d = 71 (Equation 2)
Solve the system of equations
From Equation 1:
a = 33 – 6d
Substitute this into Equation 2:
2(33 – 6d) + 13d = 71
Simplifying gives:
d = 5, a = 3
The first 5 terms are:
T1 = 3, T2 = 8, T3 = 13, T4 = 18, T5 = 23
Sum of the first 5 terms:
3 + 8 + 13 + 18 + 23 = 65
Answer: 65
