The maxim value in function y=(log base 5 (6))^sin x is a and the maxim value in function y = (log base 5 (6))^sin x is b. What is relationship between a and b?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should evaluate the maximum values of the given functions, hence, you need first find its derivatives, such that:

`(dy)/(dx) = (d((log_5 6)^sin x))/(dx)`

`(dy)/(dx) = (log_5 6)^sin x*ln(log_5 6)*(d(sin x))/(dx)`

`(dy)/(dx) = (log_5 6)^sin x*ln(log_5 6)*cosx`

You need to solve for `x` the equation `(dy)/(dx) = 0` , such that:

`(log_5 6)^sin x*ln(log_5 6)*cosx = 0`

Reducing by  `ln(log_5 6)` yields:

`(log_5 6)^sin x*cosx = 0`

Since the exponential factor `(log_5 6)^sin x > 0` yields that you need to solve the equation `cos x = 0` , such that:

`cos x = 0 => x = +-pi/2 + 2n*pi`

You need to replace `pi/2` for x in equation of the function, to evaluate its maximum value a, such that:

`a = (log_5 6)^(sin (pi/2)) => a = (log_5 6)`

Differentiating the second function with respect to `x` , yields:

`(dy)/(dx) = (d(log_5 6)^cos x)/(dx)`

`(dy)/(dx) = (log_5 6)^cos x*ln(log_5 6)*(d(cos x))/(dx)`

`(dy)/(dx) = (log_5 6)^cos x*ln(log_5 6)*(-sin x)`

You need to solve for x the equation `(dy)/(dx) = 0` , such that:

`(log_5 6)^cos x*ln(log_5 6)*(-sin x)= 0 => sin x = 0 => x = n*pi`

You need to replace `0` for `x` in equation of the function, to evaluate its maximum value `b` , such that:

`b = (log_5 6)^(cos 0) = log_5 6`

Hence, comparing the maximum values of the given functions, yields that `a = b = log_5 6.`

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