log⁡4(0.2⋅log⁡125(0.125))\log_{4}\left( 0.2 \cdot \log_{125}\left( 0.125 \right) \right)

asked by guest
on Nov 22, 2024 at 10:25 am



You asked:

Evaluate the expression: log⁡4(0.2log⁡125(0.125))\log_{4}\left( 0.2 \log_{125}\left( 0.125 \right) \right)

MathBot Answer:

log⁡4(0.2log⁡125(0.125))=log⁡(log⁡(8)5log⁡(125))+iπlog⁡(4)\log_{4}\left( 0.2 \log_{125}\left( 0.125 \right) \right) = \frac{\log{\left(\frac{\log{\left(8 \right)}}{5 \log{\left(125 \right)}} \right)} + i \pi}{\log{\left(4 \right)}}

log⁡4(0.2log⁡125(0.125))=log⁡(210⋅log⁡(1251000))=log⁡(210⋅log⁡(18))=log⁡(210⋅−⋅1log⁡(125)⋅log⁡(8))=log⁡(15⋅−⋅1log⁡(125)⋅log⁡(8))=log⁡(−15⋅1log⁡(125)⋅log⁡(8))=1log⁡(4)⋅(π⋅i+log⁡(15⋅1log⁡(125)⋅log⁡(8)))\begin{aligned}\log_{4}\left( 0.2 \log_{125}\left( 0.125 \right) \right)&=\log{\left(\frac{2}{10} \cdot \log{\left(\frac{125}{1000} \right)} \right)}\\&=\log{\left(\frac{2}{10} \cdot \log{\left(\frac{1}{8} \right)} \right)}\\&=\log{\left(\frac{2}{10} \cdot - \cdot \frac{1}{\log{\left(125 \right)}} \cdot \log{\left(8 \right)} \right)}\\&=\log{\left(\frac{1}{5} \cdot - \cdot \frac{1}{\log{\left(125 \right)}} \cdot \log{\left(8 \right)} \right)}\\&=\log{\left(- \frac{1}{5} \cdot \frac{1}{\log{\left(125 \right)}} \cdot \log{\left(8 \right)} \right)}\\&=\frac{1}{\log{\left(4 \right)}} \cdot \left(\pi \cdot i + \log{\left(\frac{1}{5} \cdot \frac{1}{\log{\left(125 \right)}} \cdot \log{\left(8 \right)} \right)}\right)\end{aligned}

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