compare-log-2-3-and-log-3-5-

Question Number 1100 by malwaan last updated on 13/Jun/15

c o m p a r e {l o g}_{2} 3 a n d {l o g}_{3} 5

Answered by 123456 last updated on 13/Jun/15

\log_{2} 3 = x \Leftrightarrow 2^{x} = 3

\log_{3} 5 = y \Leftrightarrow 3^{y} = 5

3 < 5

2^{x} < 3^{y}

\log_{2} 3 = \frac{\log 3}{\log 2} > 1

\log_{3} 5 = \frac{\log 5}{\log 3} > 1

\log 5 > \log 3 > \log 2

\log^{2} 5 > \log 3 \log 5 > \log 2 \log 5

\log 3 \log 5 > \log^{2} 3 > \log 2 \log 3

\log 2 \log 5 > \log 2 \log 3 > \log^{2} 2

\log 3 \log 5 > \log^{2} 3 > \log 2 \log 3

\log 3 \log 5 > \log 2 \log 5 > \log 2 \log 3

\dots \dots \dots . .

2^{x} < 3^{y}

x < \log_{2} 3^{y} = y \log_{2} 3 = \log_{3} {5 \log}_{2} 3

= \frac{\log 5}{\log 3} \cdot \frac{\log 3}{\log 2} = \frac{\log 5}{\log 2} = \log_{2} 5

y > \log_{3} 2^{x} = x \log_{3} 2 = \log_{2} {3 \log}_{3} 2

= \frac{\log 3}{\log 2} \cdot \frac{\log 2}{\log 3} = 1

1 < \log_{2} 3 < \log_{2} 5

1 < \log_{3} 5

\log_{2} 5 > \log_{3} 5 > 1

Commented by malwaan last updated on 14/Jun/15

w h e r e i s t h e a n s w e r ?

I s {l o g}_{2} 3 > o r < {l o g}_{3} 5 ?

Answered by prakash jain last updated on 14/Jun/15

9 > 8

3 > 2 \sqrt{2} = 2^{3 / 2}

\log_{2} 3 > 1.5 \dots (a)

25 < 27

5 < 3 \sqrt{3} = 3^{3 / 2}

\log_{3} 5 < 1.5 \dots (b)

From (a) and (b)

\log_{2} 3 > \log_{3} 5

Commented by prakash jain last updated on 15/Jun/15

\log_{20} 80 = \frac{\log 80}{\log 20} = \frac{1 + 3 \log 2}{1 + \log 2}

\log_{80} 640 = \frac{\log 640}{\log 80} = \frac{1 + 6 \log 2}{1 + 3 \log 2}

\log 2 = x

\log_{20} 80 - \log_{80} 640

[{(1 + 3 x)}^{2} - (1 + x) (1 + 6 x)] \div [(1 + x) (1 + 3 x)]

Ignoring Denominator since it does contribute to sign

2 > 1 \Rightarrow \log 2 > 0

= 1 + 6 x + 9 x^{2} - 1 - 7 x - 6 x^{2}

= 3 x^{2} - x

= x (3 x - 1)

Sign will be > 0 if x > \frac{1}{3}

x = \log 2

10 > 8

10^{1 / 3} > 2

\frac{1}{3} > \log 2 = x

Since x = \log 2 < \frac{1}{3} \Rightarrow x (3 x - 1) < 0

\Rightarrow \log_{20} 80 < \log_{80} 640

Commented by malwaan last updated on 14/Jun/15

t h a n k y o u v e r y m u c h

M r P r a k a s h

Commented by malwaan last updated on 14/Jun/15

w h a t a b o u t

{l o g}_{20} 80 a n d {l o g}_{80} 640 ?

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