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# How should I count Log1/√2(0,125)

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How should I count: Log1/√2(0,125)

Apr 26, 2015

#2
+118117
+5

Thanks Alan

This  is exactly the same as what Alan did :)

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

Apr 26, 2015

#1
+33266
+5

If logab = x then ax = b so if log1/√20.125 = x then 1/2x/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.)

.

Apr 26, 2015
#2
+118117
+5
$$\\let\;y=log_\frac{1}{\sqrt2}(0.125)\\\\ then\\\\ 0.125=\left(\frac{1}{\sqrt2}\right)^y\\\\ \frac{1}{8}=\left(\frac{1}{2^{1/2}}\right)^y\\\\ \frac{1}{2^3}=\frac{1}{2^{y/2}}\\\\ 3=\frac{y}{2}\\\\ y=6$$