I don't know how to solve this: 0,02^x - 1,02*0,02^(x+1) = -0,02 Thank you
R rudiakys New member Joined Jan 6, 2010 Messages 1 Jan 6, 2010 #1 I don't know how to solve this: 0,02^x - 1,02*0,02^(x+1) = -0,02 Thank you
D Deleted member 4993 Guest Jan 6, 2010 #2 rudiakys said: I don't know how to solve this: 0,02^x - 1,02*0,02^(x+1) = -0,02 Thank you Click to expand... 0.02x−1.02∗0.02(x+1)=−0.02\displaystyle 0.02^x - 1.02*0.02^(x+1) = - 0.020.02x−1.02∗0.02(x+1)=−0.02 0.02x−1.02∗0.02∗0.02x=−0.02\displaystyle 0.02^x - 1.02*0.02 *0.02^x = - 0.020.02x−1.02∗0.02∗0.02x=−0.02 0.02x(1−1.02∗0.02)=−0.02\displaystyle 0.02^x (1- 1.02*0.02) = - 0.020.02x(1−1.02∗0.02)=−0.02 0.02x(1−0.0204)=−0.02\displaystyle 0.02^x (1- 0.0204 ) = - 0.020.02x(1−0.0204)=−0.02 If I interpreted the problem correctly, this problem does not have solution in the real domain.
rudiakys said: I don't know how to solve this: 0,02^x - 1,02*0,02^(x+1) = -0,02 Thank you Click to expand... 0.02x−1.02∗0.02(x+1)=−0.02\displaystyle 0.02^x - 1.02*0.02^(x+1) = - 0.020.02x−1.02∗0.02(x+1)=−0.02 0.02x−1.02∗0.02∗0.02x=−0.02\displaystyle 0.02^x - 1.02*0.02 *0.02^x = - 0.020.02x−1.02∗0.02∗0.02x=−0.02 0.02x(1−1.02∗0.02)=−0.02\displaystyle 0.02^x (1- 1.02*0.02) = - 0.020.02x(1−1.02∗0.02)=−0.02 0.02x(1−0.0204)=−0.02\displaystyle 0.02^x (1- 0.0204 ) = - 0.020.02x(1−0.0204)=−0.02 If I interpreted the problem correctly, this problem does not have solution in the real domain.
