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Log base 5 of square root 17

 Nov 13, 2014

Best Answer 

 #1
avatar+118723 
+5

$$\\log_5\;\sqrt{17}\\\\
=log_5\;17^{0.5}\\\\
=0.5\times log_5\;17\\\\
=0.5\times \frac{log17}{log5}\\\\$$

 

(I used the change of base law)

 

$${\frac{{\mathtt{0.5}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{17}}\right)}{{log}_{10}\left({\mathtt{5}}\right)}} = {\mathtt{0.880\: \!187\: \!213\: \!861\: \!293\: \!8}}$$

.
 Nov 13, 2014
 #1
avatar+118723 
+5
Best Answer

$$\\log_5\;\sqrt{17}\\\\
=log_5\;17^{0.5}\\\\
=0.5\times log_5\;17\\\\
=0.5\times \frac{log17}{log5}\\\\$$

 

(I used the change of base law)

 

$${\frac{{\mathtt{0.5}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{17}}\right)}{{log}_{10}\left({\mathtt{5}}\right)}} = {\mathtt{0.880\: \!187\: \!213\: \!861\: \!293\: \!8}}$$

Melody Nov 13, 2014

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