Monday, May 13, 2013

Solve: 5^(x-1)= 5^x-5


we will denote log base 5 as log5 
5%5E%28x-1%29=+5%5Ex-5 

5%5E%28x-1%29-5%5E%28x-1%29%2B5=+5%5Ex-5%5E%28x-1%29-5%2B5 

5%5Ex-%285%5Ex%29%2F5=+5 

5%5Ex%2A%281-1%2F5%29=5 

5%5Ex=5%2F%284%2F5%29 

5%5Ex=25%2F4 

%0D%0Alog5%285%5Ex%29=log5%2825%2F4%29%0D%0A 

%0D%0Ax=log5%2825%29-log5%284%29=2-log5%284%29%0D%0A 



:)
Posted by at
Labels: , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)