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Fundamental Laws of Logarithms

If log4 5 = a...

Question

If $l o g_{4}$ 5 = a and $l o g_{5}$ 6 = b, then $l o g_{3}$ 2 is equal to

\frac{1}{2 a + 1}

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\frac{1}{2 b + 1}

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2 a b + 1

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\frac{1}{2 a b - 1}

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Solution

The correct option is D $\frac{1}{2 a b - 1}$
$a b = l o g_{4} 5. l o g_{5} 6 = l o g_{4} 6 = \frac{1}{2} l o g_{2} 6$
$a b = \frac{1}{2} (1 + l o g_{2} 3) \Rightarrow 2 a b - 1 = l o g_{2} 3$
$∴ l o g_{3} 2 = \frac{1}{2 a b - 1}$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 + a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

l o g_{4}

l o g_{5}

l o g_{3}

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