log_{7} log_{7} sqrt{7 sqrt{7 sqrt{7}}} = ? | vedclass

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Question

(C) Given expression: $\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}}$First,simplify the inner radical: $\sqrt{7 \sqrt{7 \sqrt{7}}} = (7 \cdot (7 \cdot

Vedclass · Accepted Answer

$1 - 3 \log_{7} 2$

Vedclass · Answer

(C) Given expression: $\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}}$First,simplify the inner radical: $\sqrt{7 \sqrt{7 \sqrt{7}}} = (7 \cdot (7 \cdot 7^{1/2})^{1/2})^{1/2} = (7 \cdot (7^{3/2})^{1/2})^{1/2} = (7 \cdot 7^{3/4})^{1/2} = (7^{7/4})^{1/2} = 7^{7/8}$.Now,substitute this back into the expression: $\log_{7} \log_{7} (7^{7/8})$.Using the property $\log_{b} (b^x) = x$,we get $\log_{7} (7/8)$.Using the property $\log_{b} (m/n) = \log_{b} m - \log_{b} n$,we get $\log_{7} 7 - \log_{7} 8$.Since $\log_{7} 7 = 1$ and $8 = 2^3$,the expression becomes $1 - \log_{7} (2^3)$.Using the power rule $\log_{b} (a^n) = n \log_{b} a$,we get $1 - 3 \log_{7} 2$.

$\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = ?$

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