Fun with Logarithms

Algebra Level 2

log 2 ( 1 + 1 2 ) + log 2 ( 1 + 1 3 ) + log 2 ( 1 + 1 4 ) + + log 2 ( 1 + 1 31 ) = ? \log_{2}\left(1+\frac{1}{2}\right)+\log_{2}\left(1+\frac{1}{3}\right)+\log_{2}\left(1+\frac{1}{4}\right)+\cdots+\log_{2}\left(1+\frac{1}{31}\right)=\, ?


The answer is 4.

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2 solutions

Chew-Seong Cheong
Oct 23, 2016

S = log 2 ( 1 + 1 2 ) + log 2 ( 1 + 1 3 ) + log 2 ( 1 + 1 3 ) + . . . + log 2 ( 1 + 1 31 ) = log 2 3 2 + log 2 4 3 + log 2 5 4 + . . . + log 2 32 31 = log 2 ( 3 2 × 4 3 × 5 4 × . . . × 32 31 ) = log 2 ( 3 2 × 4 3 × 5 4 × . . . × 32 31 ) = log 2 32 2 = log 2 16 = 4 \begin{aligned} S & = \log_2 \left(1+\frac 12\right) + \log_2 \left(1+\frac 13\right) + \log_2 \left(1+\frac 13\right) + ... + \log_2 \left(1+\frac 1{31}\right) \\ & = \log_2 \frac 32 + \log_2 \frac 43 + \log_2 \frac 54 + ... + \log_2 \frac {32}{31} \\ & = \log_2 \left(\frac 32 \times \frac 43 \times \frac 54 \times ... \times \frac {32}{31} \right) \\ & = \log_2 \left(\frac {\cancel 3}2 \times \frac {\cancel 4}{\cancel 3} \times \frac {\cancel 5}{\cancel 4} \times ... \times \frac {32}{\cancel{31}} \right) \\ & = \log_2 \frac {32}2 = \log_2 16 = \boxed{4} \end{aligned}

Satwik Murarka
Oct 23, 2016

log 2 ( 3 2 ) + log 2 ( 4 3 ) + . . . . . log 2 ( 32 31 ) According to properties of logarithms the above expression is equal to, log 2 ( 3 2 × 4 3 × . . . . . × 32 31 ) log 2 ( 32 2 ) log 2 16 = 4 \log_{2}(\frac{3}{2})+\log_{2}(\frac{4}{3})+.....\log_{2}(\frac{32}{31})\\ \text{According to properties of logarithms the above expression is equal to,}\\ \log_{2}(\frac{3}{2}×\frac{4}{3}×.....×\frac{32}{31})\\ \log_{2}(\frac{32}{2})\\ \log_{2}16=\boxed{4}

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