Logarithm basics

Algebra Level 2

Find the value of : $\log_{1003}(1+\frac{1}{2})+\log_{1003}(1+\frac{1}{3})+\log_{1003}(1+\frac{1}{4})+....+\log_{1003}(1+\frac{1}{2005})$

The answer is 1.

2 solutions

Yannis Wu-Yip
Dec 24, 2018

$\log_{1003}(1+\frac{1}{2})+\log_{1003}(1+\frac{1}{3})+...+\log_{1003}(1+\frac{1}{2005})$

$=\log_{1003}(\frac{3}{2})+\log_{1003}(\frac{4}{3})+...+\log_{1003}(\frac{2006}{2005})$

$=\log_{1003}((\frac{3}{2})(\frac{4}{3})...(\frac{2006}{2005}))$

$=\log_{1003}(\frac{2006}{2})$

$=\boxed{1}$

Vedant Saini
Dec 24, 2018

$\begin{aligned} \log_{1003}(1+\frac{1}{2}) + \log_{1003}(1+\frac{1}{3}) + \log_{1003}(1+\frac{1}{4}) + ... + \log_{1003}(1+\frac{1}{2005}) \\ = \log_{1003}\frac{3}{2} + \log_{1003}(\frac{4}{3}) + \log_{1003}(\frac{5}{4}) + ... + \log_{1003}(\frac{2006}{2005}) \\ = \log_{1003}3 - \log_{1003}2 + \log_{1003}4 - \log_{1003}3 + ... + \log_{1003}2006 - \log_{1003}2005 && \color{magenta} \text{Recall that: } \log_{a}\frac{m}{n} = \log_{a}m - \log_{a}n \end{aligned}$ This is precisely a telescoping series

On simplifying, we get: $\begin{aligned} \log_{1003}2006 - \log_{1003}2 \\ = \log_{1003}\frac{2006}{2} \\ =\log_{1003}1003 \\ = \boxed{1} \end{aligned}$

You could also combine the logs on the second line to the third line, getting $\log_{1003}(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot ... \cdot \frac{2006}{2005})$ . This saves one extra line of working, but both methods are just as good.

Stephen Mellor - 2 years, 5 months ago

Logarithm basics

The answer is 1.

2 solutions

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