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How should I count: Log1/√2(0,125)

 Apr 26, 2015

Best Answer 

 #2
avatar+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
avatar+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
avatar+118117 
+5
Best Answer

 

Thanks Alan

This  is exactly the same as what Alan did :)

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

Melody Apr 26, 2015

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