Question

Answer each question. BE SURE TO SHOW ALL YOUR WORK (even what you write down in your calculator). Results that are incorrect without work, will not receive any credit!

Answer 1

Explanation:

For #1, we can use what I call "the circular thing" for logs. My students always seem to remember it and they always know what I'm talking about when I tell them to use the circular thing for logs. It's unwritten but understood that the base on a common log is 10. Raising 10 to the power of whatever number is on the other side of the equals sign will be set equal to the number you're taking the log of. Yikes! In other words,

`log(1)=0` "circles" to

`10^0=1` and anything raised to the power of 0 is 1.

For #2. We will use another property of logs.

In
`5^(-x+4)=5^(x-10)` both the bases are 5's. Therefore, if we take the log base 5 of each side, they cancel each other out, leaving just the exponents set equal to each other:

`log_55^(-x+4)=log_55^(x-10)` simplifies to

-x + 4 = x - 10 (like I said, log base 5 cancels out the 5)

-2x = -14 so

x = 7

For #3, it's pretty much the same thing as #2, but on one side we have a base of 3 and the other side has a base of 27. Well, it just so happens that we can rewrite 27 in terms of 3! 3 cubed is 27, therefore,

`3^(2x-2)=27^(6x+3)` can be rewritten as

`3^(2x-2)=(3^3)^{6x+3`

The rule for power to power is that you multiply them, giving us now:

`3^(2x-2)=3^(18x+9)`

Now we'll take log base 3 of both sides, and the log base 3 and 3 will cancel, leaving the exponents only:

`log_33^(2x-2)=log_33^(18x+9)` and the log base 3 of 3 cancels out, leaving us with

2x - 2 = 18x + 9 and

-11 = 16x so

x = -.6875

For #4, we will use our calculator for some of it.

`log_3(10)=2.095903274`

That's what the right side of the equation equals, so we will write that there:

`log_3(5-x)=2.095903274`

Using the circular thing for logs again:

`3^(2.095903274)=5-x`

The left side of the equation, on your calculator, is 9.999999997 so:

9.999999997 = 5 - x and

x = -5.0000

For #5 we will use yet another property of logs, our calculator, and the circular thing again:

`2log_4(4m)+log_4(3)=2`

We can move that 2 that's in front of the log base 4 of 4m to its original exponential position, and at the same time replace what log base 4 of 3 is with what it's equal to:

`log_4(4m)^2+.7924812504=2` and

`log_4(16m^2)=1.20751875` Now we'll use the circular thing:

`4^(1.20751875)=16m^2` and

`5.333333336=16m^2` and

`.3333333335=m^2` so

m = .5774

All done! Remember all these properties and you'll be fine! All log problems are solved using these same properties over and over again.

Oh yeah, I checked them all and I promise they are all correct!!