Oh my Log

Algebra Level pending

log ⁡ 2 2 2 2 2 2 = ? \large \log_{\sqrt{\sqrt{2}}} \sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}} = \ ?

31 4 \frac {31}4 None of the others 31 8 \frac {31}8 31 32 \frac {31}{32} 32 31 \frac {32}{31}

Abu Bakar Khan by Abu Bakar Khan , 31, India

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1 solution

x = log ⁡ 2 2 2 2 2 2 = log ⁡ ( 2 1 2 ) 1 2 ( 2 ( 2 ( 2 ( 2 ( 2 ) 1 2 ) 1 2 ) 1 2 ) 1 2 ) 1 2 = log ⁡ 2 1 4 ( 2 ( 2 ( 2 ( 2 3 2 ) 1 2 ) 1 2 ) 1 2 ) 1 2 = log ⁡ 2 1 4 ( 2 ( 2 ( 2 7 4 ) 1 2 ) 1 2 ) 1 2 = log ⁡ 2 1 4 ( 2 ( 2 15 8 ) 1 2 ) 1 2 = log ⁡ 2 1 4 ( 2 31 16 ) 1 2 = log ⁡ 2 1 4 ( 2 31 32 ) = 31 8 \begin{aligned} x & = \log_{\sqrt{\sqrt{2}}} \sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}} \\ & = \log_{\left(2^\frac 12\right)^\frac 12} \left(2\left(2 \left(2 \left(2\left(2 \right)^\frac 12 \right)^\frac 12\right)^\frac 12\right)^\frac 12 \right)^\frac 12 \\ & = \log_{2^\frac 14} \left(2\left(2 \left(2 \left(2^\frac 32 \right)^\frac 12\right)^\frac 12\right)^\frac 12 \right)^\frac 12 \\ & = \log_{2^\frac 14} \left(2\left(2 \left(2^{\frac 74} \right)^\frac 12\right)^\frac 12 \right)^\frac 12 \\ & = \log_{2^\frac 14} \left(2\left(2^{\frac {15}8} \right)^\frac 12 \right)^\frac 12 \\ & = \log_{2^\frac 14} \left(2^{\frac {31}{16}} \right)^\frac 12 \\ & = \log_{2^\frac 14} \left(2^{\frac {31}{32}} \right) \\ & = \boxed{\dfrac {31}8} \end{aligned}

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