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Question
Find k if the following function is continuous at the points indicated against them:
`f(x) = (45^x - 9^x - 5^x + 1)/((k^x - 1)(3^x - 1))` for x ≠ 0
= `2/3` for x = 0, at x = 0
Solution
f is continuous at x = 0
∴ `lim_(x→) "f"(x)` = f(0)
∴ `lim_(x→0) ((45)^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1)) = 2/3`
∴ `lim_(x→0) (9^x*5^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1)) = 2/3`
∴ `lim_(x→0) (9^x(5^x - 1) - 1(5^x - 1))/(("k"^x - 1)(3^x - 1)) = 2/3`
∴ `lim_(x→0) ((5^x - 1)(9^x - 1))/(("k"^x - 1)(3^x - 1)) = 2/3`
∴ `lim_(x→0) (((5^x - 1)(9^x - 1))/x^2)/((("k"^x - 1)(3^x - 1))/x^2) = 2/3 ...[(because x -> 0"," therefore x ≠ 0 therefore x^2 ≠ 0),("Divide Numerator and Denominator by" x^2)]`
∴ `(lim_(x→0)((5^x - 1)/x)((9^x - 1)/x))/(lim_(x→0)(("k"^x - 1)/x)((3^x - 1)/x)) = 2/3`
∴ `(log5 * log9)/(log"k" * log 3) = 2/3` `[because lim_(x→0) ("a"^x - 1)/x = log "a"]`
∴ `(log 5 * log(3)^2)/(log"k"*log3) = 2/3`
∴ `(2 xx log5 xx log3)/(log"k" xx log3) = 2/3`
∴ `log5/log "k" = 1/3`
∴ 3log 5 = log k
∴ log(5)3 = log k
∴ (5)3 = k
∴ k = 125
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