Suppose that f is an expotential function f(x)=AB^x, where A and B>0 are constants.?

Question

a) Explain why f(5)/f(3)=F(10)/f(8).
b)Find a number a such that f(4)/f(1)=f(a)/f(-9).
c) Explain why (f(2)/f(-1)^3)=f(10)/f(1)
d) Show that the value of f(x+h)/f(x) does not depend on x.
e) The value of the expression in part (d) does depend on h. Show that the value of the expression (ln(f(x+h)/f(x)))/h does not depend on h.

I am completely lost and was looking for some help...anybody understand this stuff?

Yeah, (c) was written wrong it's:

Explain why (f(2)/f(-1))^3=f(10)/f(1)

Answer 1

So, f(x) = AB^x. I'll assume it's only B raised to the x power, although it doesn't matter really.

a) f(5) = AB^5. f(3) = AB^3. f(5)/f(3) = AB^5/AB^3 = B^(5-3) = B^2.

f(10) = AB^10. f(8) = AB^8. f(10)/f(8) = AB^10/AB^8 = B^(10-8) = B^2.

So all that really matters is the difference between the the number on top and the number on the bottom.

b) Since 4-1 is 3, you want a number on top that is 3 more than -9. That number would be -6.

c) Doesn't work, think you must have written it down wrong. On the right you'd have B^(10-1) = B^9. On the left you've got B^(2-(-1)) = B^(2+1) = B^3. So I think the left side should read: (f(2)/f(-1))^3, right? The ^3 has to be outside the parentheses, because (B^3)^3 = B^(3*3) = B^9.

d) f(x+h) = AB^(x + h). f(x) = AB^x.

f(x+h)/f(x) = AB^(x+h)/AB^x = AB^(x+h-x) = AB^h

Since x drops out, the expression doesn't depend on x.

e) We just showed that f(x+h)/f(x) = AB^h.

So ln [f(x+h)/f(x)] = ln AB^h = h ln AB, by the laws of logarithms.

Then ln [f(x+h)/f(x)]/h = h ln AB/h = ln AB.

The h drops out, so the result does not depend on h.

Answer 2

First, note that f(x)/f(y) = AB^x / AB^y = B^x/B^y = B^(x-y).
Therefore f(a)/f(b) = f(c)/f(d) is equivalent to saying a-b = c-d. Knowing this, the answer to all five questions practically writes itself...

a) 5-3 = 10-8, so the equality follows from what I just said.

b) This means 4-1 = a - (-9); so a = -6.

c) This only works if the exponent is applied to the whole fraction. In this case you have
(f(2)/f(-1))³ = [B^(2-(-1))]³ = (B³)³ = B^9 = RHS.

d) Again using the above result, this is equal to B^(x+h - x) = B^h. Nuff said.

e) This equals ln(B^h)/h = h*lnB/h = lnB.

Answer 3

a)
a^5 / a^3 = a^(5-3) = a^2
a^10 / a^8 = a^(10-8) = a^2

b) Look at a) and notice you have to make sure that:
4-1 = a+9 -> a = -6

c)
(f(2)/f(-1))³ = (a^2/a^-1)³ = (a^3)³ = a^9 = f(10)/f(1)

d)
f(x+h) / f(x) = a^(x+h)-x = a^h

e)
ln(f(x+h)/f(x))/h = (ln a^h) / h = (h ln a) / h = ln a

Answer 4

a) Just substitute the values

=>f(5)=AB^5, similarly f(3) and so on....
=>(AB^5)/(AB^3)=AB^2

Also (AB^10)/(AB^8)=AB^2

=>f(5)/f(3)=f(10)/f(8)

b)This is similar as the first one

=>(AB^4)/AB=(AB^a)/(AB^(-9))
=>AB^3=AB^(9+a)
=>3=9+a
=>a=-6

c)f(2)/(f(-1))^3=((AB^2)/((AB^-1))^3
=(AB^3)^3=AB^9

Also, f(10)/f(1)=(AB^10)/AB=AB^9

Therefore, (f(2)/f(-1)^3)=f(10)/f(1)

d)f(x+h)/f(x)

=>AB^(x+h)/AB^x
=>AB^(x+h-x)=AB^h

e)(ln(f(x+h)/f(x)))/h

Solve from inside to outside
=>f(x+h)/f(x)=AB^h...........already calculated in (d)
Now, ln(a) means log (to the base e)(a)
=>ln(AB^h)=h ln(AB)
(h ln(AB))/h=ln(AB).........There is no h in the ultimate expression therefore whatever be the value of h the expression remains the same, hence independant of h.

Answer 5

(f(x)+i*g(x))*(f(y)+i*g(y)) =f(x)f(y)-g(x)g(y) +i*(f(x)g(y)+g(x)f(y)) =f(x+y)+ig(x+y) so (f(x)+i*g(x))*(f(y)+i*g(y)) =f(x+y)+ig(x+y) and so |f(x)+i*g(x)|*|f(y)+i*g(y)| =|f(x+y)+ig(x+y)| Now define h(x)=ln|f(x)+i*g(x)| Then h(x)+h(y)=h(x+y) h is non-end so h(x)=cx for some consistent c. this implies ln|f(x)+i*g(x)|=cx subsequently |f(x)+i*g(x)|=exp(cx) meaning f(x)^2+g(x)^2=exp(2cx) i'm uncertain a thank you to apply f'(0)=0 to end c=0 for the needed result.

Answer 6

a.f(5)/f(3)=AB^5/AB^3=B^2
f(10)/f(8)=AB^10/AB^8=B^@
so AB^5/AB^3=AB^10/AB^8
=>f(5)/f(3)=f(10)/f(8)

b.f(4)/f(1)=B^3
f(a)/f(-9)=AB^a/AB^-9
=B^(a+9)
B(a+9)=B^3
a+9=3
a=-6

c.{f(2)/f(-1)}^3={AB^2/AB^-1}3=B^9
f(10)/f(1)=AB^10/AB^1=B^9

d.f(x+h)/f(x)=AB^(x+h)/AB^x
=AB^(x+h-x)=AB^h so independent of x

e.lnf(x+h)/f(x)}/h=hlnAB/h
=lnAB.so independentof h