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Singapore Math Olympiad

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

Check the Answer


Answer: 3.5<x<4

Singapore Mathematical Olympiad

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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Categories
Singapore Math Olympiad

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

Check the Answer


Answer: 36

Singapore Mathematical Olympiad, 2008

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)

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