+0  
 
0
296
3
avatar

Please solve and explain me like I was 5
3^(2n-1) < 10^ (n-1)    where n is natural

Guest Feb 28, 2015

Best Answer 

 #2
avatar+92781 
+5

thanks anon, that was a good start. :)

I am going to use     log base 10     because      $$log_{10}10=1$$

 

$$\\3^ {2n-1} < 10^ {n-1 }\\\\
Log_{10}3^ {2n-1} (2n-1)Log_{10}3<(n-1)Log_{10} 10\\\\
Log_{10}3(2n-1)<(n-1)\\\\
2nLog_{10}3-Log_{10}3 2nLog_{10}3-n n(2Log_{10}3-1)

 

$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.045\: \!757\: \!490\: \!560\: \!675\: \!1}}$$    

 

I am going to divide by this number and since it is negative I will have to turn the sign around.

 

$$\\n>\frac{log_{10}(3)-1}{2log_{10}(3)-1}$$

 

$${\frac{\left({log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}} = {\mathtt{11.427\: \!172\: \!663\: \!391\: \!417\: \!6}}$$

 

$$n\ge 12 \qquad where \;\;n\in N$$ 

Melody  Feb 28, 2015
 #1
avatar
+5

here is my helpm you incorporate the log on the left and right and continue like so

Guest Feb 28, 2015
 #2
avatar+92781 
+5
Best Answer

thanks anon, that was a good start. :)

I am going to use     log base 10     because      $$log_{10}10=1$$

 

$$\\3^ {2n-1} < 10^ {n-1 }\\\\
Log_{10}3^ {2n-1} (2n-1)Log_{10}3<(n-1)Log_{10} 10\\\\
Log_{10}3(2n-1)<(n-1)\\\\
2nLog_{10}3-Log_{10}3 2nLog_{10}3-n n(2Log_{10}3-1)

 

$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.045\: \!757\: \!490\: \!560\: \!675\: \!1}}$$    

 

I am going to divide by this number and since it is negative I will have to turn the sign around.

 

$$\\n>\frac{log_{10}(3)-1}{2log_{10}(3)-1}$$

 

$${\frac{\left({log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}} = {\mathtt{11.427\: \!172\: \!663\: \!391\: \!417\: \!6}}$$

 

$$n\ge 12 \qquad where \;\;n\in N$$ 

Melody  Feb 28, 2015
 #3
avatar+92781 
0

I wanted to give you thumbs up anon but it won't let me. 

There seems to be some problems with the system tonight.   

Melody  Feb 28, 2015

12 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.