Use the quadratic formula to solve the equations . you will get the answers
Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.
Consider this: (2)^a negative power = (1/2)^the same power but positive.
So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.
What I just said in that paragraph was: log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.
Now let's look at the problem:
log₂(x-1) + log(base 1/2) (x-2) = log₂(x)
Subtract log₂(x) from each side:
log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0
Subtract log(base 1/2) (x-2) from each side:
log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.
The left side is the same as log₂[ (x-1)/x ]
==> The right side is the same as +log₂(x-2)
Now you have: log₂[ (x-1)/x ] = +log₂(x-2)
And that ugly [ log to the base of 1/2 ] is gone.
Take the antilog of each side:
(x-1)/x = x-2
Multiply each side by 'x' : x - 1 = x² - 2x
Subtract (x-1) from each side:
x² - 2x - (x-1) = 0
x² - 3x + 1 = 0
Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .
I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.
There,now. Doesn't that feel better.
The answer is 19 6×13 It’s 78-97 you get 19
Answer:
option-C
Step-by-step explanation:
We are given row-echelon form
So, we can write equation from matrix as
Let's assume
now, we can find x and y
so, we get solutions as
where t can be any real numbers
