Question

Let f(x) = logb(9x5 − 8). for what value of b is f ′(1) = 9

Answer 1

To find the value of b for which f'(1) = 9, we need to differentiate the function f(x) = logb(9x^5 - 8) with respect to x.

Step-by-step explanation:

To find the value of b for which f'(1) = 9, we need to differentiate the function f(x) = logb(9x^5 - 8) with respect to x. Applying the chain rule, we have:

f'(x) = (1/(9x^5 - 8)) * (45x^4) * (1/ln(b))

Substituting x = 1 and f'(1) = 9 into the derivative expression, we get:

9 = (1/(9 - 8)) * (45) * (1/ln(b))

Simplifying further, we find:

9 = 45 / ln(b)

To isolate b, we can cross multiply and then take the reciprocal:

ln(b) = 45 / 9 = 5

b = e^5 ≈ 148.41316