Write a rational equation that relates the desired percentage to the amount of % acid solution that needs to be added to liter o
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Answer:
Step-by-step explanation:
For a. we start by dividing both sides by 200:
In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:
Applying the power rule for logs lets us now bring down the x in front of the ln:
x * ln(1.05) = ln(1.885)
Now we can divide both sides by ln(1.05) to solve for x:
Do this on your calculator to find that
x = 12.99294297
For b. we will first apply the rule for "undoing" the addition of logs by multipllying:
Simplifying gives you
Applying the power rule allows us to bring down the 3 in front of the ln:
3 * ln(x) = 5
Now we can divide both sides by 3 to get
Take the inverse ln by raising each side to e:
The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that
x = 5.29449005
For c. begin by dividing both sides by 20 to get:
"Undo" that e by taking the ln of both sides:
When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:
-.6931471806 = .1x
x = -6.931471806
Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:
Now we will bring over the -8 by adding:
We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let
When we do that, we can rewrite the polynomial as
This factors very nicely into u = 4 and u = 2
But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:
For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:
Now that the bases are the same, we can say that
x = 2
For the second solution, we will raise the 2 on the right to a power of 1 to get:
Now that the bases are the same, we can say that
x = 1
