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P1*(V1)^n=P2*(v2)^n      

P1=100 , P2=200

V2=V1*5

find n ?

 Jun 12, 2014

Best Answer 

 #1
avatar+33661 
+5

$$100*v_1^n=200*(5v_1)^n\\\\
100*v_1^n=200*5^n*v_1^n\\\\
1=2*5^n\\\\
$ Take logs\\
$$\ln{1}=\ln{2}+n*\ln{5}\\\\

0=\ln{2}+n*\ln{5}\\\\
n=-\frac{\ln{2}}{\ln{5}}$$

$${\mathtt{n}} = {\mathtt{\,-\,}}{\frac{{ln}{\left({\mathtt{2}}\right)}}{{ln}{\left({\mathtt{5}}\right)}}} \Rightarrow {\mathtt{n}} = -{\mathtt{0.430\: \!676\: \!558\: \!073\: \!393\: \!1}}$$

.
 Jun 12, 2014
 #1
avatar+33661 
+5
Best Answer

$$100*v_1^n=200*(5v_1)^n\\\\
100*v_1^n=200*5^n*v_1^n\\\\
1=2*5^n\\\\
$ Take logs\\
$$\ln{1}=\ln{2}+n*\ln{5}\\\\

0=\ln{2}+n*\ln{5}\\\\
n=-\frac{\ln{2}}{\ln{5}}$$

$${\mathtt{n}} = {\mathtt{\,-\,}}{\frac{{ln}{\left({\mathtt{2}}\right)}}{{ln}{\left({\mathtt{5}}\right)}}} \Rightarrow {\mathtt{n}} = -{\mathtt{0.430\: \!676\: \!558\: \!073\: \!393\: \!1}}$$

Alan Jun 12, 2014

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