Logarithm Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Logarithm topic of quantitative aptitude
(a) 1.5625
(b) 4.2500
(c) 2.5625
(d) 1.2500
The correct answers to the above question in:
Answer: (a)
We have, y = $x^{logx}$
Taking log on both sides log
y = log $(x^{logx})$
= logx. logx = $(1.25)^2$ = 1.5625
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If a = $\text"log"_{24} 12, b = \text"log"_{36}$ 24, C = $\text"log"_{48}$ 36. Then 1 + abc is equal to
a) 2ab
b) 2ac
c) 2abc
d) 2bc
e) None of these
Answer »Answer: (d)
abc = ${\text"log" 12}/{\text"log" 24} . {\text"log" 24}/{\text"log" 36} . {\text"log" 36}/{\text"log" 48} = {\text"log" 12}/{\text"log" 48}$
∴ 1 + abc = ${\text"log" 48 + \text"log"12}/{\text"log" 48} = {\text"log" (48.12)}/{\text"log" 48}$
= ${\text"log"24}^2/{\text"log" 48} = 2.{\text"log" 24}/{\text"log" 48}$ = 2bc
Question : 2
If ${\text"log"_a} b = 1/2, {\text"log"_b}$ c = $1/3$ and ${\text"log"_c}$ a = $k/5$, then the value of k is
a) 30
b) 25
c) 20
d) 35
e) None of these
Answer »Answer: (a)
${\text"log"_a} b = {1/2}, {\text"log"_b}c= {1/3}$, and ${\text"log"_c} a = k/5$
⇒${\text"log" b}/{\text"log" a} = {1/2}, {\text"log" c}/{\text"log" b} = {1/3}, {\text"log" a}/{\text"log" c} = k/5$
⇒${1/2} × {1/3} × {k/5}$ = 1⇒k = 30
Question : 3
What is the solution of the equation x $log_{10} ({10}/3) + log_{10} 3 = log_{10} (2 + 3^x)$ + x ?
a) 1
b) 10
c) 3
d) 0
Answer »Answer: (d)
$x log_{10}({10}/3) + log_{10} 3 = log_{10} (2 + 3^x) + x$
$xlog_{10} 10 - x log_{10} 3 + log_{10} 3 = log_{10} (2 + 3^x)$ + x
x - $log_{10^3^x} + log_{10^3} = log_{10} (2 + 3^x)$ + x
$log_{10[3/{3^x}]} log_{10} (2 + 3^x)$
⇒$3^{1–x} = 2 + 3^x$
⇒$3^{1–x} –3^x = 3^1 – 3^0$
x = 0
Question : 4
What is the value of $(\text"log"_{1/2} 2) (\text"log"_{1/3}3) (\text"log"_{1/4}4).....(\text"log"_{1/{1000}}1000)$
a) 1 or – 1
b) 1
c) 0
d) – 1
e) None of these
Answer »Answer: (d)
$(\text"log"_{1/2}2)(\text"log"_{1/3}3)(\text"log"_{1/4}4).......(\text"log"_{1/{1000}}1000)$
=$({\text"log" 2}/{\text"log"{1/2}})({\text"log" 3}/{\text"log"{1/3}})({\text"log" 4}/{\text"log"{1/4}})......({\text"log" 1000}/{\text"log"{1/{1000}}})$ $(∵ \text"log"_b a = {\text"log" a}/{\text"log" b})$
= $({\text"log" 2}/{- \text"log" 2})({\text"log" 3}/{-\text"log" 3})({\text"log" 4}/{-\text"log" 4}) ...... ({\text"log" 1000}/{\text"log" 1000})$
= (-1) × (-1) × (-1) × .....× (-1)
(∵ number of terms is odd) = -1
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