Math Is Fun Forum

Mizter

Guest

Logarithm

Hi,

I just have a question concerning logarithms. It should be very easy to solve because I think I am just missing out on a basic property.

I found this on a website and I'm not sure how the guy got to the first step. I'm assuming its some sort of property that I have not learned. Can anyone help me out?

>Given: log_a(x) = c and log_b(x) = d.
>Find:  log_ab(x), the log of x in base (a*b), in terms of c and d.

>I'll post my solution in a day or two, but I'm curious if someone
>else comes up with something more elegant.

   x^(1/c) = a     x^(1/d) = b <--- how do i get to this step?

   ab = x^(1/c + 1/d) = x^[ (c+d) / (cd) ]

   log_ab(ab) = 1 = (c+d)/(cd) log_ab(x)

Thanks for your help

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Logarithm

log_a(x) = c

Divide by c:
(1/c)log_a(x) = 1

Bring the coefficient inside the log:
log_a(x^(1/c)) = 1

Raise both sides to the power of a:
x^(1/c) = a

The other one is exactly the same method with different letters.

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Why did the vector cross the road?
It wanted to be normal.

help please

Guest

Re: Logarithm

how is (1/c) log[base a] x = 1  equal to log[base a] (x^1/c) = 1 ?

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 51,348

Re: Logarithm

help please,
this is a simple property in logarithms.
If

doesn't it follow that

?

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