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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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## Trigonometry Problem from SMO, 2008 | Problem No.17

Try this beautiful Problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

## Problem – Trigonometry (SMO Test)

Find the value of $(log_{\sqrt 2}(cos 20^\circ) + log_{\sqrt 2} (cos 40^\circ) + log_{\sqrt 2}(cos 80^\circ))^2$

• 32
• 15
• 36
• 20

### Key Concepts

Trigonometry

Log Function

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

This one is a very simple. We can start from here :

As all are in the function of log with $\sqrt 2$ as base so we can take it as common such that

$log_{\sqrt 2}(cos 20 ^\circ . cos 40 ^\circ . cos 80^\circ)$

Now as you can see we dont know the exact value of $cos 20^\circ$ or $cos 40^\circ$ or $cos 80^\circ$ values.

But theres a formula that we can use which is

cosA.cos B = $\frac {1}{2} (cos (A+B) + cos (A-B))$

Now try apply this formula in the above expression and try to solve………

Now those who did not get the answer yet try this:

If we apply the formula in the expression mentioned in the last hint :

$log_{\sqrt 2}(cos 20 ^\circ . \frac {1}{2}(cos 120 ^\circ . cos 40^\circ)$

$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{2} cos 40 ^\circ . cos 20^\circ)$

$log_{\sqrt 2}(-\frac {1}{4} cos 20 ^\circ +\frac {1}{4} (cos 60 ^\circ + cos 20^\circ)$

We need to do the rest of the calculation.Try to do that …………………..

Continue from the last hint:

$log_{\sqrt 2} \frac {1}{8} = – \frac {log_{2} 8}{log_{2}(2^{\frac {1}{2}})} = -6$

So squaring this answer = $(-6)^2 = 36$ ……………………..(Answer)