SOLUTION: What is the value of R in this equation: (1-R^-48).02083 = R - 1

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Question 117743: What is the value of R in this equation:
(1-R^-48).02083 = R - 1
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%281-R%5E-48%29%28.02083%29=+R-1

%281-1%2FR%5E48%29+%28.02083%29+=+R-1

+%28.02083%29+-+%281%2FR%5E48%29+%28.02083%29+=+R-1....multiply both sides by R%5E48

+%28.02083%29R%5E48+-+%281%2FR%5E48%29+R%5E48%28.02083%29+=+R%5E48%28R-1%29....

+%28.02083%29R%5E48+-+%28.02083%29+=+R%5E48%28R-1%29....move +%28.02083%29R%5E48+ to the right

+-+%28.02083%29+=+R%5E48%28R-1%29+-+%28.02083%29R%5E48+....factor out common R%5E48

+-+%28.02083%29+=+R%5E48%28R+-+1+-.02083%29....

+-+%28.02083%29+=+R%5E48%28R+-+1.02083%29.... take a look at the left and the right side of the equation; in order to get -.02083 on the right side, R have to be equal 1
so, your answer is:
R=1
check:
+-+%28.02083%29+=+R%5E48%28R+-+1.02083%29....
+-+%28.02083%29+=+1%5E48%281-+1.02083%29
+-+%28.02083%29+=+1%28-.02083%29
+-+%28.02083%29+=+%28-.02083%29