# Solving equation with logs and variables

• Nov 27th 2011, 09:17 AM
benny92000
Solving equation with logs and variables
Problem: F(x) = 4^(x) for all real values of x. If p>1 and q>1, then f^(-1)(p)f^(-1)(q)=?

The notation above for f^(-1) means inverse function.

The answer is log(base 4) p x log(base 4)q

I first attempted to solve for the inverse function by switching y and x and then solving for y. This did not get me log (base4) x, which the book says is the inverse of the exponential function.

(Nerd)
• Nov 27th 2011, 12:42 PM
Siron
Re: Solving equation with logs and variables
• Nov 27th 2011, 05:03 PM
benny92000
Re: Solving equation with logs and variables
I didn't come to the right answer, but I have y=4^(x) I switched x and y to solve for the inverse. x=4^(y) I took the log of both sides logx= ylog4. Then I got logx/log4. I then plugged in p and q and fiddled with it from there to no avail.
• Nov 27th 2011, 05:49 PM
skeeter
Re: Solving equation with logs and variables
I assume again ...

$\displaystyle f(x) = 4^x$ (you used a capital F)

you should know that the inverse of an exponential function is a log function, but here is the derivation anyway ...

$\displaystyle y = 4^x$

$\displaystyle x = 4^y$

$\displaystyle \log_4(x) = \log_4(4^y)$

$\displaystyle \log_4(x) = y$

so ...

$\displaystyle f^{-1}(x) = \log_4(x)$

$\displaystyle f^{-1}(p) \cdot f^{-1}(q) = \log_4(p) \cdot \log_4(q)$
• Nov 28th 2011, 05:11 AM
HallsofIvy
Re: Solving equation with logs and variables
Quote:

Originally Posted by benny92000
I didn't come to the right answer, but I have y=4^(x) I switched x and y to solve for the inverse. x=4^(y) I took the log of both sides logx= ylog4. Then I got logx/log4. I then plugged in p and q and fiddled with it from there to no avail.

Which logarithm did you use? $\displaystyle a^x$ gives a different function for every different a, so its inverse $\displaystyle log_a(x)$ is a different function for every a. If you had used $\displaystyle log_4(x)$, Since $\displaystyle 4= 4^1$, log_4(4)= 1 so your "log x/log 4" is just $\displaystyle log_4(x)$. For any other logarithm, it is true that $\displaystyle log_a()x)/log_a(4)= log_4(x)$. That's a property of logarithms worth knowing:
$\displaystyle \frac{log_a(x)}{log_a(b)}= log_b(x)$
• Nov 28th 2011, 10:20 AM
benny92000
Re: Solving equation with logs and variables
Change of base rule. I am aware of it. I see how that applies here now. Thanks!