An algebra problem by Abu Bakar Khan

Algebra Level 2

log ⁡ 8 ( i i 2 + i + i 2432 + i 3616 ) = ? \large \log_8 \left(i^{i^2} + i + i^{2432} + i^{3616}\right) = \ ?

Notation: i = − 1 i = \sqrt {-1} is the imaginary unit .

None of the rest 2 -1 1 3 -2

Abu Bakar Khan by Abu Bakar Khan , 31, India

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1 solution

log ⁡ 8 ( i i 2 + i + i 2432 + i 3616 ) = log ⁡ 8 ( i − 1 + i + ( i 4 ) 608 + ( i 4 ) 904 ) = log ⁡ 8 ( 1 i + i + ( 1 ) 608 + ( 1 ) 904 ) = log ⁡ 8 ( − i + i + 1 + 1 ) = log ⁡ 8 ( 2 ) = log ⁡ 8 ( 8 3 ) = 1 3 \begin{aligned} \log_8 \left(i^{i^2} + i + i^{2432} + i^{3616} \right) & = \log_8 \left(i^{-1} + i + \left(i^4\right)^{608} + \left(i^4\right)^{904} \right) \\ & = \log_8 \left(\frac 1i + i + \left(1\right)^{608} + \left(1\right)^{904} \right) \\ & = \log_8 \left(-i + i + 1 + 1 \right) \\ & = \log_8 (2) = \log_8 (\sqrt[3]{8}) = \frac 13 \end{aligned}

Therefore, the answer is None of the rest. \boxed{\text{None of the rest.}}

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