First slide

Trigonometric Identities

Question

If $\log_{10} 5 = a and \log_{10} 3 = b$ then

Moderate

A

$\log_{30} 8 = \frac{3 (1 - a)}{b + 1}$
B

$\log_{30} 8 = \frac{3 (1 - a)}{b + 1}$
C

$\log_{243} 32 = \frac{1 - a}{b}$
D

none of these

Solution

$\begin{matrix} (1) \log_{30} 8 = \frac{3 \log_{10} 2}{\log_{10} 5 + \log_{10} 3 + \log_{10} 2} \\ = \frac{3 (1 - \log_{10} 5)}{\log_{10} 5 + \log_{10} 3 + \log_{10} 2} \\ = \frac{3 (1 - \log_{10} 5)}{1 + \log_{10} 3} = \frac{3 (1 - a)}{1 + b} \end{matrix}$
$\begin{matrix} (2) \log_{40} 15 = \frac{\log_{10} 15}{\log_{10} 40} \\ = \frac{\log_{10} 3 + \log_{10} 5}{\log_{10} 5 + 3 [1 - \log_{10} 5]} = \frac{a + b}{3 - 2 a} \end{matrix}$
$(3) \log_{243} 32 = \frac{\log_{10} 2}{\log_{10} 3} = \frac{1 - \log_{10} 5}{\log_{10} 3} = \frac{1 - a}{b}$

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