How should I count Log1/√2(0,125)

 

Thanks Alan

This  is exactly the same as what Alan did :)

 $$\\let\;y=log_\frac{1}{\sqrt2}(0.125)\\\\
then\\\\
0.125=\left(\frac{1}{\sqrt2}\right)^y\\\\
\frac{1}{8}=\left(\frac{1}{2^{1/2}}\right)^y\\\\
\frac{1}{2^3}=\frac{1}{2^{y/2}}\\\\
3=\frac{y}{2}\\\\
y=6$$

If log_ab = x then a^x = b so if log_1/√20.125 = x then 1/2^x/2 =0.125  

Now 0.125 = 1/8 = 1/2³, so we must have x/2 = 3 or x = 6

 

Check

$${{log}}_{{\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{2}}}}}}}{\left({\mathtt{0.125}}\right)} = {\mathtt{6.000\: \!000\: \!000\: \!000\: \!001}}$$

(The 1 at the end is just numerical error.)

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