Quantitative Aptitude
LOGARITHM MCQs
Logarithms
(a) Since loga a = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.
log5 512 = \(\frac{\log512}{\log5}\)
= \(\frac{\log2^{9}}{\log(\frac{10}{2})}\)
= \(\frac{9\log2}{\log10-\log2}\)
= \(\frac{(9\times0.3010)}{1-0.3010}\)
= \(\frac{2.709}{0.699}\)
= \(\frac{2709}{699}\)
=3.876
\(\frac{\log8}{\log8} = \frac{\log8^{\frac{1}{2}}}{\log8}=\frac{\frac{1}{2}\log8}{\log8}=\frac{1}{2}\)
log \(\frac{a}{b}\) +log \(\frac{b}{a}\) = log (a + b)
log (a + b) = log \(\log\left(\frac{a}{b}\times\frac{b}{a}\right)\) = log 1.
o, a + b = 1.
log10 \(\left(\frac{1}{70}\right)\) = log10 1 - log10 70
= - log10 (7 x 10)
= - (log10 7 + log10 10)
= - (a + 1).
\(\log_{2}10=\frac{1}{\log_{10}2}=\frac{1}{0.310}=\frac{10000}{3010}=\frac{1000}{301}\)
log10 80 = log10 (8 x 10)
= log10 8 + log10 10
= log10 (23 ) + 1
= 3 log10 2 + 1
= (3 x 0.3010) + 1
= 1.9030.
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10
log10 [5 (5x + 1)] = log10 [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x = 3.
Given expression= log60 3 + log60 4 + log60 5
= log60 (3 x 4 x 5)
= log60 60
= 1.
log (264)= 64 x log 2
= (64 x 0.30103)
= 19.26592
Its characteristic is 19.
Hence, then number of digits in 264 is 20.