Answer:
Step-by-step explanation:
For a. we start by dividing both sides by 200:
![(1.05)^x=1.885](https://tex.z-dn.net/?f=%281.05%29%5Ex%3D1.885)
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:
![ln(1.05)^x=ln(1.885)](https://tex.z-dn.net/?f=ln%281.05%29%5Ex%3Dln%281.885%29)
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:
![x=\frac{ln(1.885)}{ln(1.05)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bln%281.885%29%7D%7Bln%281.05%29%7D)
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:
![ln(x*x^2)=5](https://tex.z-dn.net/?f=ln%28x%2Ax%5E2%29%3D5)
Simplifying gives you
![ln(x^3)=5](https://tex.z-dn.net/?f=ln%28x%5E3%29%3D5)
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
![ln(x)=\frac{5}{3}](https://tex.z-dn.net/?f=ln%28x%29%3D%5Cfrac%7B5%7D%7B3%7D)
Take the inverse ln by raising each side to e:
![e^{ln(x)}=e^{\frac{5}{3}}](https://tex.z-dn.net/?f=e%5E%7Bln%28x%29%7D%3De%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
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:
![\frac{1}{2}=e^{.1x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3De%5E%7B.1x%7D)
"Undo" that e by taking the ln of both sides:
![ln(.5)=ln(e^{.1x})](https://tex.z-dn.net/?f=ln%28.5%29%3Dln%28e%5E%7B.1x%7D%29)
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:
![(2^2)^x-6(2)^x=-8](https://tex.z-dn.net/?f=%282%5E2%29%5Ex-6%282%29%5Ex%3D-8)
Now we will bring over the -8 by adding:
![(2^2)^x-6(2)^x+8=0](https://tex.z-dn.net/?f=%282%5E2%29%5Ex-6%282%29%5Ex%2B8%3D0)
We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let ![u=2^x](https://tex.z-dn.net/?f=u%3D2%5Ex)
When we do that, we can rewrite the polynomial as
![u^2-6u+8=0](https://tex.z-dn.net/?f=u%5E2-6u%2B8%3D0)
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:
![2^x=4,2^x=2](https://tex.z-dn.net/?f=2%5Ex%3D4%2C2%5Ex%3D2)
For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:
![2^2=2^x](https://tex.z-dn.net/?f=2%5E2%3D2%5Ex)
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:
![2^x=2^1](https://tex.z-dn.net/?f=2%5Ex%3D2%5E1)
Now that the bases are the same, we can say that
x = 1