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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). 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

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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