If f(x)= 9^x then prove that f(m+n+p) = f(m).f(n).f(p)​

f(x)=9^(x)

Plugging in m+n+p=x, we have

f(m+n+p) = 9^(m+n+p)

Using the properties of exponential a^(b+c) = a^b*a^c, we have

f(m+n+p) = 9^(m)*9^(n)*9^(p)

f(m+n+p) = f(m)*f(n)*f(p)

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752