Name: Logarithms in Five Minutes
Uploaded: 2019-07-19
Duration: 5 min 43 s
Description: Logarithms are a very interesting and useful concept used in maths, physics and many more areas. Let's gather all the important information about them in a f...

Question

If

\log_{12} 18 = x, \log_{24} 54 = y

, then show that

x y + 5 (x - y) = 1

.

Accepted Answer

Given: $x = \log_{12} 18$

We know that, $\log_{a} b = \frac{\log b}{\log a}$

$\Rightarrow x = \frac{\log_{2} 18}{\log_{2} 12} = \frac{\log_{2} (2 \times 9)}{\log_{2} (3 \times 4)}$

$\Rightarrow x = \frac{\log_{2} 2 + \log_{2} 9}{\log_{2} 3 + \log_{2} 4} = \frac{\log_{2} 2 + \log_{2} 3^{2}}{\log_{2} 3 + \log_{2} 2^{2}} = \frac{\log_{2} 2 + 2 \log_{2} 3}{\log_{2} 3 + 2 \log_{2} 2}$

$\Rightarrow x = \frac{1 + 2 \log 3}{2 + \log 3}$

Given: $y = \log_{24} 54 = \frac{\log_{2} 54}{\log_{2} 24} = \frac{\log_{2} (6 \times 9)}{\log_{2} (8 \times 3)}$

$\Rightarrow y = \frac{\log_{2} 6 + \log_{2} 9}{\log_{2} 8 + \log_{2} 3} = \frac{\log_{2} (2 \times 3) + \log_{2} 3^{2}}{\log_{2} 2^{3} + \log_{2} 3} = \frac{\log_{2} 2 + \log_{2} 3 + 2 \log_{2} 3}{3 \log_{2} 2 + \log_{2} 3}$

$\Rightarrow y = \frac{1 + 3 \log 3}{3 + \log 3}$

Consider L.H.S: $x y + 5 (x - y)$

$\Rightarrow (\frac{2 \log 3 + 1}{2 + \log 3}) (\frac{3 \log 3 + 1}{3 + \log 3}) + 5 ((\frac{2 \log 3 + 1}{2 + \log 3}) - (\frac{3 \log 3 + 1}{3 + \log 3}))$

$\Rightarrow \frac{6 {(\log 3)}^{2} + 2 \log 3 + 3 \log 3 + 1}{6 + 5 \log 3 + {(\log 3)}^{2}} + 5 (\frac{1 - {(\log 3)}^{2}}{6 + 5 \log 3 + {(\log 3)}^{2}})$

$\Rightarrow \frac{6 {(\log 3)}^{2} + 2 \log 3 + 3 \log 3 + 1 + 5 - 5 {(\log 3)}^{2}}{6 + 5 \log 3 + {(\log 3)}^{2}}$

$\Rightarrow \frac{6 + 5 \log 3 + {(\log 3)}^{2}}{6 + 5 \log 3 + {(\log 3)}^{2}} = 1$

$L . H . S = R . H . S$

Hence, proved.

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