This EduRev document offers 15 Multiple Choice Questions (MCQs) from the topic Exponents & Logarithm (Level - 2). These questions are of Level - 2 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If x + 1/x = 3, then the value of is Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If – 7 = 0, then two values of x are Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:Find the value of. Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:Which of the following is true?
Explanation
log_{11} 1650 = log_{11} (11 × 150) = log_{11} 11 + log_{11} 150 = 1 + log_{11} 150
Now, as 112 = 121 and as 121 < 150, we have log_{11} 150 > 2.
Thus, log_{11} 1650 > 1 + 2.
Or, log_{11} 1650 > 3
log_{13} 1950 = log_{13} (13 × 150) = log_{13} 13 + log_{13} 150 = 1 + log_{13} 150
Now, as 13^{2} = 169 and as 169 > 150, we have log_{13} 150 < 2.
Thus, log_{13} 1950 < 1 + 2 or log_{13} 1950 < 3.
⇒ log_{13} 1950 < log_{11} 1650
Hence, answer option a is correct.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If x = log24 + log35 + log46 + log57 + log68 + log79, then which of the following is true?
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If1/3 log_{3} M + 3 log_{3} N = 1 + log0.008 5, then
Explanation
The given equation boils down to the following.
Hence, answer option b is correct.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If then find the value of a^{a} b^{b} c^{c}. Explanation
Let
Thus, loga = (b - c)d; logb = (c - a)d; logc = d(a - b)
Now, let a^{a}b^{b}c^{c} = f
Or, log(a^{a}b^{b}c^{c}) = logf
Or, a loga + b logb + c logc = logf
Plugging in the valued of loga, logb & logc in terms of d in the above equation,
we get
a (b - c)d + b(c - a)d + c(a - b)d = logf
or 0 = log f
or, f = 1
∴ a^{a}b^{b}c^{c} = 1
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If log_{10} x - log_{10}= 2 log_{x} 10, then x is equal to Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If log (a + c), log (c - a), log (a - 2b + c) are in A.P, then
Explanation
log (a + c), log (c - a), log (a - 2b + c) are in A.P.
log (c - a) - log (a + c) = log (a - 2b + c) - log (c - a)
c^{2} + a^{2} - 2ac = a^{2} + c^{2} + 2ac - 2ab - 2bc
4ac = 2ab + 2bc
So, a, b, c are in H. P.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:Arrange the following in descending order: A = 18^{22}, B = 22^{18}, C = 21^{17}
Explanation
A = 18^{22}, B = 22^{18}, C = 21^{17}
We know that a^{b} > b^{a}, where a < b and a, b ≥ 3.
Thus, 18^{22} > 22^{18}
Moreover, 18^{22} > 22^{18} > 21^{17}
Thus, the required sequence is ABC. Hence, option (a) is correct.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:The sum of the series x^{2} + + ..., |x| < 1, is Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:Find the value of x in log (x - 2) + log (x - 1) = log 20 - 1.
Explanation
The given equation is equivalent to:
log (x - 2)(x - 1) = log 20 - log 10
log (x - 2)(x - 1) = log 2
(x - 2)(x - 1) - 2 = 0
x2 - 3x + 2 - 2 = 0
x(x - 3) = 0
x = 0 or x = 3
x = 0 is not admissible since logarithm of a negative number (when used in the given equation) is not defined.
∴ x = 3
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:If log_{12 }27 = a and log9 16 = b, then find log_{8} 108.
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:equals Explanation
Thus, the expression becomes
log_{xyz} xy + log_{xyz} yz + log_{xyz} xz
= logxyz (x^{2}y^{2}z^{2}) = logxyz(xyz)^{2} = 2log_{xyz} xyz = 2
Question for Practice Questions Level 2: Exponents & Logarithm - 2
Try yourself:Let u = (log_{2 }x)^{2} - 6 log2 x + 12, where x is a real number. Then, the equation x^{u} = 256 has
Explanation
u = (log_{2} x)^{2} - 6 log_{2} x + 12 & x^{u }= 256 = 2^{8} = 4^{4} = 16^{2} = 256^{1}
If x = 2 ⇒ u = (log_{2} 2)^{2} - 6 log_{2} 2 + 12
u = 1 - 6 + 12 = 7 (Not possible)
If x = 4 ⇒ u = (log_{2} 4)^{2} - 6 log_{2} 4 + 12
= (2 log_{2} 2)^{2} - 12 log_{2} 2 + 12
= 4 - 12 + 12 = 4 (Possible)
If x = 16 ⇒ u = (log_{2} 16)2 - 6 log_{2} 16 + 12
u = 16 - 24 + 12
u = 4 (Not possible)
If x = 256 ⇒ u = (8 log_{2} 2)2 - 48 log_{2} 2 + 12
= 64 - 48 + 12
u = 28 (Not possible)
x can have exactly one solution.