# Thread: Solving equation with logs and variables

1. ## 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.

3. ## 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.

4. ## Re: Solving equation with logs and variables

I assume again ...

$f(x) = 4^x$ (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 ...

$y = 4^x$

$x = 4^y$

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

$\log_4(x) = y$

so ...

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

$f^{-1}(p) \cdot f^{-1}(q) = \log_4(p) \cdot \log_4(q)$

5. ## Re: Solving equation with logs and variables

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? $a^x$ gives a different function for every different a, so its inverse $log_a(x)$ is a different function for every a. If you had used $log_4(x)$, Since $4= 4^1$, log_4(4)= 1 so your "log x/log 4" is just $log_4(x)$. For any other logarithm, it is true that $log_a()x)/log_a(4)= log_4(x)$. That's a property of logarithms worth knowing:
$\frac{log_a(x)}{log_a(b)}= log_b(x)$

6. ## Re: Solving equation with logs and variables

Change of base rule. I am aware of it. I see how that applies here now. Thanks!