Ok man this is how you go
in question 2 p5
suppose
I/V V/V lg(I/I) lg(V/V)
2 4+/- 0.1
3 7+/- 0.1
you can simply calculate values of Log I example log 2 =0.301 write it in the column
lg(I/I)
for log V you calculate value Log V example log 4= 0.602 write it in the column under lg(V/V)
absolute errors in lg (V/V):
error in V/V is +/- 0.1 thus absolute error in lg (V/V) would be *
take the first value in the group V/V which is 4
(log 4) - (log 3.9) = 0.01 to 1 sig fig
or
(log 4.1)- (log 4) = 0.01 to 1 sig fig
note that 3.9 or 4.1 value comes from subtracting or adding 0.1 from actual value which was 4 see column above of V/V
both ways are correct
same for the second value 7 +/- 0.1
log 7.1 - log 7.0
or
log 7 - log 6.9
another example
100 +/- 20
you do for absolute errors
log 100 -log 80
or
log 120 -log 100
just remember to either add or subtract the error from actual value
tricky
if its square of distance
suppose d/cm = 2 cm+/- 0.1
the errors in "d square world be"
(2.1 square ) - 2.0 square
or
2.0 square - 1.9 square
this would give you error
in this case d square = 4+/-0.4
hope it solves it all for every one
