Answer:
![x =\pm \sqrt{26}](https://tex.z-dn.net/?f=%20x%20%3D%5Cpm%20%5Csqrt%7B26%7D)
Step-by-step explanation:
<u>Given </u><u>:</u><u>-</u><u> </u>
And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,
<u>Equation</u><u> </u><u>:</u><u>-</u><u> </u>
<u>In </u><u>general</u><u> </u><u>:</u><u>-</u><u> </u>
- If we have a logarithmic equation as ,
Then this can be written as ,
In a similar way we can write the given equation as ,
- Now also we know that
Therefore , the equation becomes ,
<u>Hence</u><u> the</u><u> </u><u>Solution</u><u> </u><u>of </u><u>the</u><u> given</u><u> equation</u><u> is</u><u> </u><u>±</u><u>√</u><u>2</u><u>6</u><u>.</u>
Answer:
200 miles
Step-by-step explanation:
Let y(t) represent the level of water in inches at time t in hours. Then we are given ...
y'(t) = k√(y(t)) . . . . for some proportionality constant k
y(0) = 30
y(1) = 29
We observe that a function of the form
y(t) = a(t - b)²
will have a derivative that is proportional to y:
y'(t) = 2a(t -b)
We can find the constants "a" and "b" from the given boundary conditions.
At t=0
30 = a(0 -b)²
a = 30/b²
At t=1
29 = a(1 - b)² . . . . . . . . . substitute for t
29 = 30(1 - b)²/b² . . . . . substitute for a
29/30 = (1/b -1)² . . . . . . divide by 30
1 -√(29/30) = 1/b . . . . . . square root, then add 1 (positive root yields extraneous solution)
b = 30 +√870 . . . . . . . . simplify
The value of b is the time it takes for the height of water in the tank to become 0. It is 30+√870 hours ≈ 59 hours 29 minutes 45 seconds