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# Logarithm Problem From SMO, 2011 | Problem 7

Try this beautiful Logarithm Problem From Singapore Mathematics Olympiad, SMO, 2011 (Problem 7). You may use sequential hints to solve the problem.

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

But try the problem first…

Source

Challenges and thrills – Pre – College Mathematics

## Try with Hints

First hint

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 ………………………………..

Second Hint

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

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 …………………..

Final Step

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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