Calculus help! Need help with 2 derivative problems!

find the derivative of each function.

a) f(x) = 20^8x



b) g(x)= log(sub 5) 25x

a) f '(x) = 8*ln(20)*20^(8x)

b) To do this one without the equation for deriving logs you would do:
y = log5_(25x)
Rewrite
5^y = 25x
Derive
dy/dx*ln(5)*5^y = 25
dy/dx = 25/( ln(5)*5^y )
Plug in log5_(25x) for y
dy/dx = 25/( ln(5)*5^(log5_(25x)) )
Simplify
g'(x) = 25/( ln(5)*25x )
Simplify even more (didn't want to skip too far ahead with simplifying)
g'(x) = 1/( ln(5)*x )

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I have checked these answers and am 100% sure of their correctness.
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a.) f(x) = 20⁸ⁿ

If f(x) = a·x^(b·x), then d/dx [f(x)] = ln(a)·b·a^(b·x)

d/dx [f(x)] = ln(20)·(8)·(20)⁸ⁿ
→ 8·ln(20)·25600000000ⁿ
Solution: d/dx [20⁸ⁿ] = 8·ln(20)·25600000000ⁿ

b.) g(x) = log5(25x)

If f(x) = logb(a·x), then d/dx [f(x)] = logb(e)/x

d/dx [g(x)] = log5(e)/x
Solution: d/dx [log5(25x)] = log5(e)/x

a)
LHS: d ln(f)/dx = 1/f df/dx

RHS: d ln[20^8x)]/dx = d/dx { 8x.ln(20) } = 8ln(20)

df/dx = 20^8x.8ln(20)

b)
g(x) = log (sub5) 25x = (1/ln5) ln(25x)

dg/dx = (1/ln5) (25/25x) = (1/x)log(sub5) e or x/ln5

a)2^(16 x+3) 5^(8 x) log(20)

b)log(sub 5) 25x=(log(25 x))/(log(5))

derivative 1/(x log(5))

a) f'(x) = 8*ln(20)*20^8x
b) g(x) = ln(25x)/ln(5)
g'(x) = 25/ln(5) * 1/(25x)
g'(x) = 1/(ln(5)*25x)