• Dec 4th 2008, 07:24 AM
shirkdeio
What is the relationship between logarithms and radicals?
Here's my question: what is the relationship between radicals and logarithms? How are they related?

(My original question is below, but the answer occurred to me after I posted it)

I have a simple question--why is the inverse of an exponential function a logarithmic function instead of a radical function?

If $F(x) = 5^x$, wouldn't the inverse be $x = 5^y$, and couldn't that be written $\sqrt[5]{x}$? Why must it be written $log_5x$
• Dec 4th 2008, 07:52 AM
running-gag
Quote:

Originally Posted by shirkdeio
Here's my question: what is the relationship between radicals and logarithms? How are they related?

(My original question is below, but the answer occurred to me after I posted it)

I have a simple question--why is the inverse of an exponential function a logarithmic function instead of a radical function?

If $F(x) = 5^x$, wouldn't the inverse be $x = 5^y$, and couldn't that be written $\sqrt[5]{x}$? Why must it be written $log_5x$

Hi

If $y = 5^x$ then $x = log_5 \,y$

If $y = x^5$ then $x = \sqrt[5] y$
• Dec 4th 2008, 08:24 AM
shirkdeio
Quote:

Originally Posted by running-gag
Hi

If $y = 5^x$ then $x = log_5 \,y$

If $y = x^5$ then $x = \sqrt[5] y$

Thanks. I don't know why I had trouble with that--logarithms confuse me.

But is $y = 5^x$ the inverse of $\sqrt[x]{y} = 5$? If so, what is the relationship/difference between the radical and $y = log_5{x}$?
• Dec 4th 2008, 08:38 AM
running-gag
Quote:

Originally Posted by shirkdeio
Thanks. I don't know why I had trouble with that--logarithms confuse me.

But is $y = 5^x$ the inverse of $\sqrt[x]{y} = 5$? If so, what is the relationship/difference between the radical and $y = log_5{x}$?

The inverse function of y=f(x) is $x = f^{-1}(y)$
You have to express x function of y

If $y = 5^x$ then $x = log_5\,y$

$\sqrt[x]{y} = 5$ is also true but it is not in the form $x = f^{-1}(y)$

It is the same as per the example :
If $y = 5x$ then $x = \frac{y}{5}$

$\frac{y}{x} = 5$ is also true but it is not in the form $x = f^{-1}(y)$
• Dec 4th 2008, 10:57 AM
shirkdeio
Thank you. I think I better understand logarithms now. I appreciate the help!