1. ## exponent equation

Just a simple question. How do we find R if it is given by the equation

[(100-R)^0.5]+[(400-R)^0.5]=20

I have tried taking the natural logs on both sides of the equation, but it does not return the correct answer, which is 93.75.

2. Originally Posted by kenobings
Just a simple question. How do we find R if it is given by the equation

[(100-R)^0.5]+[(400-R)^0.5]=20

I have tried taking the natural logs on both sides of the equation, but it does not return the correct answer, which is 93.75.

Hello,

to get rid of the square roots you must square both sides:

[(100-R)^0.5]+[(400-R)^0.5]=20

100-R + 2(100-R)^0.5*(400-R)^0.5 + 400-R = 400

(100-R)^0.5*(400-R)^0.5 = R - 50 Square again:

(100-R)(400-R) = Rē - 100R + 2500

40.000 - 500R + Rē = Rē - 100R + 2500

37.500 = 400R ==> R = 93.75

3. thanks very much for the help.. any idea why taking natural logs on both sides wouldn't work?

4. Originally Posted by kenobings
thanks very much for the help.. any idea why taking natural logs on both sides wouldn't work?
we do not know how to evaluate log(x + y) unless we know exactly how to combine x and y into one number, which in this case we don't. also, taking logs is more trouble than it's worth here, since if we decide to separate the bases so we can take logs, we would have different bases, so we'd end up simplifying one and leaving (or possibly complicating) the other base

note, here your bases are (100-R) and (400-R)