Reason: givens values of log(2) and log(5) are rounded (both base 7)Just to add something, I wanna say calculating the logarithms on the calculator wont work for instances like ii), where the value is actually 2.01 but the answer will be 2.02 as that is the method the markers want you to use.
Is there any point to the 0.36 and 0.83 values?
Yeah he substituted them inYou have to use the numbers 2 and 5 to get 0.4 and 50.
(Let log(x)=log7(x))
For (i), log(0.4) = log(2/5) = log(2) - log(5) = 0.36 - 0.83 = -0.47.
For (ii) log(50) = log(25 * 2) = log(52) + log(2) = 2log(5) + log(2) = 2(0.83) + 0.36 = 2.02.