Tag: logaritm

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.