# Thread: Finding inputs for an equation r(x)=2ln1.8(1.8^x)

1. ## Finding inputs for an equation r(x)=2ln1.8(1.8^x)

I have two questions. The first is rather simple: in any situation where the outputs of a function, f(x), are given, should I naturally think to find the inverse of the equation? For this problem, r(x)= 2ln1.8(1.8^x), I'm given two outputs, r(x)=9.4, r(x)=30. I have several problems like this one on my study sheet, and I am wondering if my natural instinct should be to just find the inverse of the equation. I've heard other students say they just replace r(x) with the output and solve the equation that way, but that seems like it would take slightly longer. I can't actually solve for the inverse or solve for the output in either scenario.

Here's what I've been doing on r(x)=2ln1.8(1.8^x).

1) I use the logarithm property log AB = log A + log B. So, r(x)=2ln1.8 + ln(1.8^x)
2) I move the x. r(x)=2ln1.8 + xln1.8
3) I raise both natural logs to the power of e. r(x)=2*e^ln1.8 + x*e^ln1.8
4) e to the natural log cancels. r(x)=2*1.8 + x*1.8
5) r(x)=3.6+1.8x

Now, at this point can I just invert the equation? y=3.6+1.8x => x=3.6+1.8y => x-3.6=1.8y => x-2=y

I know this can't be right because I've tried plugging in some values and they don't agree with the normal equation. Did I make a mistake as I went through the process of breaking down the original equation?

2. ## Re: Finding inputs for an equation r(x)=2ln1.8(1.8^x)

You made a mistake with step (3) - you can't just raise the ln functions to the power of e; you must do the same to both sides of the equation:

e^(r(x)) = e^(2 ln 1.8 + x ln1.8) = e^(2ln1.8) * e^(x ln 1.8)

I think it's easier to find the inverse like this:

r = 2 ln (1.8 * 1.8^x) = 2 ln (1.8)^(x+1) = 2(x+1) ln(1.8)
Divide through by 2 ln(1.8): r/(2 ln (1.8)) = x+1
Rearrange to get x by itself: x= r/(2 ln(1.8)) - 1

So if r = 9.4 you have x = 9.4/(2 ln(1.8)) - 1 = 6.996

To check that this is correct, try x=6.996 in the original equation: r = 2 ln(1.8 * 1.8^6.996) = 2 ln(1.8^7.996) = 2 ln(109.9) = 9.4

3. ## Re: Finding inputs for an equation r(x)=2ln1.8(1.8^x)

Originally Posted by ebaines

r = 2 ln (1.8 * 1.8^x) = 2 ln (1.8)^(x+1) = 2(x+1) ln(1.8)
Thanks for your insight, Ebaines. But how did you manipulate the equation to remove the 1.8^x and replace it with (x+1)?

4. ## Re: Finding inputs for an equation r(x)=2ln1.8(1.8^x)

$\displaystyle a^m\times a^n = a^{m+n}$

5. ## Re: Finding inputs for an equation r(x)=2ln1.8(1.8^x)

Oh yeah! There's always a 1 exponent over any number...