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# Logarithm Problem From SMO, 2011 | Problem 7

Try this beautiful Logarithm Problem From Singapore Mathematics Olympiad, SMO, 2011 (Problem 7).

## Logarithm Problem From SMO

1. Let $x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}$+$\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}$+$\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}$ Which of the following statements
is true?
• 1.5<x<2
• 2<x<2.5
• 2.5<x<3
• 3<x<3.5
• 3.5<x<4

### Key Concepts

log function

Logarithmic

Inverse Exponentiation

Challenges and thrills - Pre - College Mathematics

## Try with Hints

If you got stuck in this problem we can start from here:

$x=\frac {1}{\log_{\frac {1}{3}} \frac {1}{2}}$+$\frac {1}{\log_{\frac {1}{5}} \frac {1}{4}}$+$\frac {1}{\log _{\frac {1}{7}} \frac{1}{8}}$

If we refer too the basic properties of log we can find ,

x=$\frac{\log \left(\frac{1}{3}\right)}{\log \left(\frac{1}{2}\right)}$+$\frac{\log \left(\frac{1}{5}\right)}{\log \left(\frac{1}{4}\right)}$+$\frac{\log \left(\frac{1}{7}\right)}{\log \left(\frac{1}{8}\right)}$=$\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}$

Try the rest ......................................

$\frac{-\log 3}{-\log 2}+\frac{-\log 5}{-\log 4}+\frac{-\log 7}{-\log 8}$

so we can find

$\frac {\log 3+ \log 5^{\frac {1}{2}}}+ \log 7^{\frac {1}{3}}{log 2}$

= $\frac {\log \sqrt {45} + log 7^{\frac {1}{3}}}{log 2}$ < $\frac {\log \sqrt {65} + log 8^{\frac {1}{3}}}{log 2}$

= $\frac{3 \log 2+\log 2}{\log 2}=4$

Try the rest .......................

Now let's say ,

2x = $2 \frac {log 3 + log 5^{\frac {1}{2}}+ log 7^{\frac {1}{3}}}{log 2}$

=

$\begin{array}{l} \frac{\log (9 \times 5)+\log \left(49^{\frac{1}{3}}\right)}{\log 2}>\frac{\log \left(45 \times 27^{\frac{1}{3}}\right)}{\log 2} = \ \frac{\log (45 \times 3)}{\log 2}>\frac{\log (128)}{\log 2}=7 \end{array}$

so x is greater than 3.5.

3.5 <x<4 is the correct answer.

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