arthham

JEE Advanced 2018 · Relations and Functions (Class 12)

The value of ((log⁡29)2)1log⁡2(log⁡29)×(7)1log⁡47\left((\log_2 9)^2\right)^{\frac{1}{\log_2(\log_2 9)}} \times (\sqrt{7})^{\frac{1}{\log_4 7}} is ________.

Answer: 88

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Written solution

The value of ((log₂ 9)²)^(1/(log₂(log₂ 9))) × (√7)^(1/(log₄ 7)) is ________.

((log⁡29)2)1log⁡2(log⁡29)\left((\log_2 9)^2\right)^{\frac{1}{\log_2(\log_2 9)}}

× (7)1log⁡47\times\ (\sqrt{7})^{\frac{1}{\log_4 7}}

To find: value of the whole expression\text{To find: value of the whole expression}

log⁡ba=1log⁡ab\log_b a = \dfrac{1}{\log_a b} blog⁡bx=xb^{\log_b x} = x 1. simplify factor 11.\ \text{simplify factor }1 2. simplify factor 22.\ \text{simplify factor }2 3. multiply3.\ \text{multiply} a=log⁡29a = \log_2 9 F1=(a2)1log⁡2a=a2log⁡2aF_1 = (a^2)^{\frac{1}{\log_2 a}} = a^{\frac{2}{\log_2 a}} 2log⁡2a=2log⁡a2=log⁡a4\dfrac{2}{\log_2 a} = 2\log_a 2 = \log_a 4 F1=alog⁡a4=4F_1 = a^{\log_a 4} = 4 F2=(7)1log⁡47F_2 = (\sqrt7)^{\frac{1}{\log_4 7}} 1log⁡47=log⁡74\dfrac{1}{\log_4 7} = \log_7 4 F2=(7)log⁡74F_2 = (\sqrt7)^{\log_7 4} F2=(71/2)log⁡74=712log⁡74=41/2=2F_2 = \left(7^{1/2}\right)^{\log_7 4} = 7^{\frac12\log_7 4} = 4^{1/2} = 2 F1×F2=4×2=8F_1 \times F_2 = 4 \times 2 = 8 8\boxed{8}

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