Question 57
Given that $$\frac{1}{\log _{a+1}(91)} + \frac{1}{\log _{b+1}(91)} = 1$$. If a, b are integers, find the value of a+b.
Solution
$$\log _{91} (a+1) + \frac{1}{\log _{b+1}(91)} = 1$$ can be written as
$$\log _{91} (a+1) + \log _{91}(b+1) = 1$$
==> (a+1)(b+1) = 91
==> (a,b) can be (90,0), (12,6), (6,12) or (0,90) (Since a+1, b+1 > 0).
==> But a and b cannot be 0 as the base of the logarithm cannot be 1.
==> a+b = 18
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