Answer:
Step-by-step explanation:
For a. we start by dividing both sides by 200:
[tex](1.05)^x=1.885[/tex]
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:
[tex]ln(1.05)^x=ln(1.885)[/tex]
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:
[tex]x=\frac{ln(1.885)}{ln(1.05)}[/tex]
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:
[tex]ln(x*x^2)=5[/tex]
Simplifying gives you
[tex]ln(x^3)=5[/tex]
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
[tex]ln(x)=\frac{5}{3}[/tex]
Take the inverse ln by raising each side to e:
[tex]e^{ln(x)}=e^{\frac{5}{3}}[/tex]
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:
[tex]\frac{1}{2}=e^{.1x}[/tex]
"Undo" that e by taking the ln of both sides:
[tex]ln(.5)=ln(e^{.1x})[/tex]
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:
[tex](2^2)^x-6(2)^x=-8[/tex]
Now we will bring over the -8 by adding:
[tex](2^2)^x-6(2)^x+8=0[/tex]
We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let [tex]u=2^x[/tex]
When we do that, we can rewrite the polynomial as
[tex]u^2-6u+8=0[/tex]
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:
[tex]2^x=4,2^x=2[/tex]
For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:
[tex]2^2=2^x[/tex]
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:
[tex]2^x=2^1[/tex]
Now that the bases are the same, we can say that
x = 1
