SOLUTION: I have tried and tried on this problem and just cant get it right. I even had a tutor online help me and they got it wrong too... this is the problem ... Solve the system of equ

Question 631098: I have tried and tried on this problem and just cant get it right. I even had a tutor online help me and they got it wrong too... this is the problem ...
Solve the system of equations. Round approximate values to the nearest ten-thousandth.
y = log2x
y = x−5
Found 3 solutions by richwmiller, solver91311, KMST:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
How do you know the tutor got it wrong?
Apparently, you are not telling all.
There are actually two real solutions.
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

There is no algebraic solution to this equation. You have to use numerical methods.

I used a graphing program and then ran some numbers through the function in Excel. I was able to come up with two intervals: [0.000005,0.000006] and [6,6.1] where the function changes sign from one end of the interval to the other. The Intermediate value theorem says there has to be a root on each of those intervals.

Then I used the Newton-Raphson method:

I first defined the function as

Then Newton-Raphson says

and then a closer approximation can be obtained from:

And you can repeat the process as often as you like to get the desired precision.

The derivative of is:

You can run a calculator as well as I can. Enjoy.

John

My calculator said it, I believe it, that settles it

Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
Maybe something in the problem got "lost in translation."
For example, I wonder if you meant
for the first function,
or maybe it was
The answer would depend on what that first function is.
In either of those cases, the log function and the y=x-5 function graphs will cross at two points, like this:
, so you would have 2 solutions (which I will show you further down).

It could also be that the problem is neither of those, and something in the problem is irretrievably "lost in translation."

It could also be that the "officially correct answer" that you were given is wrong, or that it needs to be entered in a specific format. After seeing my youngest child through (online) high school courses, I would not be surprised. Sometimes the official answer was wrong, and sometimes the computer would accept only (8,3), but not (8, 3) or the other way around.

If it is ,
for , and .
So but they are very close.
For a smaller x, , and , with but they and very close.
That means that the answer for x must be a number in between, that I would round to 0.00001 if I was allowed, but it rounds to 0.0000 when rounded to the nearest ten-thousandth.
The answer for y is -5.0000 when rounded to the nearest ten-thousandth, because for , for , and for everything in between, the values of and rounds to -5.0000.
That solution would be expressed as (0.0000, -5.0000).
On the other end, I found that seemed to be the closest I could get, with
and
=1.0853 (rounded).
That would give you the (x, y) pair (6.0853, 1.0853).

If it was , with or (3, 2) is a nice exact answer. Another solution is with which rounded gives you the pair (0.0319, -4.9681).
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