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