If f(u)=16^u, show that f(a) x f(b) = f(a+b)?

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2007-03-18 22:40:23 · 5 answers · asked by Ricardo 1 in Science & Mathematics ➔ Mathematics

if f(u)=16^u,so
f(a)=16^a , f(b)=16^b
f(a) x f(b)=16^a x 16^b=16^(a+b)=f(a+b)

2007-03-18 23:00:36 · answer #1 · answered by shiva 3 · 0⤊ 0⤋

I don't think you can show that, if the f of u is 16 to the u, then the f of a times the f of b should be 16 times 16 to the a times b.

2007-03-18 22:52:06 · answer #2 · answered by peteryoung144 6 · 0⤊ 1⤋

f(a) x f(b) = 16^(a) x 16^(b) = 16^(a + b) = f^(a + b)

2007-03-19 00:25:20 · answer #3 · answered by Como 7 · 0⤊ 0⤋

f(a) x f(b)=(16^a) x (16^b)=16^(a+b)=f(a+b)

2007-03-18 23:00:45 · answer #4 · answered by Anonymous · 0⤊ 0⤋

f(u) = 16^u,so

f(a) = 16^a , and f(b) = 16^b

f(a) x f(b) = 16^a x 16^b

= 16^(a+b) = f(a+b)

2007-03-18 23:20:52 · answer #5 · answered by Kinu Sharma 2 · 0⤊ 0⤋