10-31-2012
|
#14
|
|
Senior Member
Joined: | Dec 2011 |
Posts: | 2,511 |
|
Quote:
Originally Posted by dexternjack
I hurried through mine w/o thinking all the way. I have a minor in Math so I tend to over-think it and solve too fast. I made a minor error in translating, here it is...
2^(x+1) = 5^(3x)
log2^(x+1) = log 5^(3x)
(x+1)log 2 = (3x)log 5
xlog2 + log2 = 3xlog5
xlog2 - 3xlog5 = -log2
x(log2 - 3log5) = -log2
x = -log2 / (log2 - 3log5) -this is the form your teacher was asking for...
if you were to solve for x, then you would get .167
It has been a very long time since I have done these, but I am pretty confident this is correct.
|
It is.
|
|
|