Question

How do we convert this function into a logarithmic one?
P(x) = 100(1 + 0.1)^x

Answer 1

Well we could use magic but lets not :P
Anyways!
rewrite to be

p(x) / 100 = (1+0.1)^x
Remember that
If y = 2^x
x = log(2) y

Answer 2

I can isolate to

P(x) /100 = 100*(1.1)^x / 100
P(x) /100 = (1.1)^x

but then I dont understand how a function can be
P(x)/100

If I just change it to y
I get
y /100 = (1.1)^x

and then I use the log conversion
y = b^x ↔ x = log_b y

and get
x = log_1.1 (y/100)

but now how do I get back to a function ?
like how do I isolate y again?

I think I need something like...
P(x) = ???

If I try to use something like the quotient property
x = log_1.1 (y) - log_1.1 (100)
then maybe
x +log_1.1 (100) = log_1.1 (y) - log_1.1 (100) + log_1.1 (100)
x +log_1.1 (100) = log_1.1 (y)

umm then somehow try to convert log_1.1 (y) into something that lets me isolate y
if y = b^x ↔ x = log_b y

then I guess
y =1.1^(x +log_1.1 (100) )

And so
P(x) = 1.1^(x +log_1.1 (100) )

That all looks very wrong.. or is that actually right in some way?
am I confused enough?

What am I missing here.

Answer 3

Okayy. I need some time to read this :P The question is simply rewriting to log function right?

Answer 4

Actually,maybe it is just that I don't understand what they mean by "convert your exponential function P(x) into a logarithmic one"
Does this mean that while P(x) originally would return the product after applying an exponent (x), that now they want to return the exponent after giving it a product so x=a target balance for example and it returns a number of years? Is that what they mean by a logarithmic function?

Answer 5

Looking at the question, this should be fine.