It is given that y = A(10^bx), where A and b are constants. The straight line graph obtained when lgy is plotted against x passes through the points (0.5, 2.2) and (1.0, 3.7).
- (i)Find the value of A and of b.[5]
- (ii)Using your values of A and b, find the value of y when x = 0.6,[2]
- (iii)Using your values of A and b, find the value of x when y = 600.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] lg y = lg A + bx
- [M1] equating gradient to b
- [A1] b = 3
- [M1] Use of substitution into one of the following 2.2 = lg A + 0.5b OR 3.7 = lg A + b
- [A1] A = 5, 5.01 or 10^0.7
- [M1] correct statement using their A and b correctly in either equation or using lg y = 3x + 0.7
- [A1] y = 316, 315 or 10^2.5
- [M1] correct statement using their A and b correctly in either equation or using lg 600 = 3x + 0.7
- [A1] x = 0.693
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