Logs, Roots And Sevens

Algebra Level 2

log 7 log 7 7 7 7 = ? \large \log_7 \log_7 \sqrt{7\sqrt{7\sqrt 7}} = \, ?

0 0 1 1 3 log 2 7 3\log_2 7 3 log 7 2 3\log_7 2 1 3 log 2 7 1- 3\log_2 7 1 3 log 7 2 1- 3\log_7 2 Cannot be determined

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1 solution

Chew-Seong Cheong
Jun 29, 2016

Relevant wiki: Properties of Logarithms - Basic

log 7 log 7 7 7 7 = log 7 log 7 ( 7 ( 7 ( 7 ) 1 2 ) 1 2 ) 1 2 = log 7 log 7 ( 7 ( 7 3 2 ) 1 2 ) 1 2 = log 7 log 7 ( 7 7 3 4 ) 1 2 = log 7 log 7 ( 7 7 4 ) 1 2 = log 7 log 7 7 7 8 = log 7 7 8 = log 7 7 log 7 8 = 1 log 7 2 3 = 1 3 log 7 2 \begin{aligned} \log_7 \log_7 \sqrt{7\sqrt{7\sqrt 7}} & = \log_7 \log_7 \left( 7 \left(7 \left(7\right)^\frac 12 \right)^\frac 12 \right)^\frac 12 = \log_7 \log_7 \left( 7 \left(7^\frac 32 \right)^\frac 12 \right)^\frac 12 = \log_7 \log_7 \left( 7 \cdot 7^\frac 34 \right)^\frac 12 = \log_7 \log_7 \left(7^\frac 74 \right)^\frac 12 \\ & = \log_7 \log_7 7^\frac 78 = \log_7 \frac 78 = \log_7 7 - \log_7 8 = 1 - \log_7 2^3 = \boxed{1-3\log_7 2} \end{aligned}

From where that 3/2 came

Vedant Sarnobat - 4 years, 7 months ago

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7 ( 7 1 2 ) = 7 1 7 1 2 = 7 1 + 1 2 = 7 3 2 7\left(7^\frac 12\right) = 7^1\cdot 7^\frac 12 = 7^{1+\frac 12} = 7^\frac 32

Chew-Seong Cheong - 4 years, 7 months ago

Law of exponent: 7^1 x 7^1/2 = 7 ^ (1+ 1/2) = 7^3/2

Brendix Emata - 4 years, 7 months ago

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