Logs - can anyone help?

Question

I'll use the letter v to symbolize a down arrow... not sure how else to type logs..... ^ means up.

1) Rewrite as a single logarithm: 3 logv4^x - logv4^y

2) Which _expression is = to logv8[5(x-4)]
a) logv8^5x - logv8^20
b) 5 logv8^(x-4)
c) logv8^x
d) logv8^5 + logv8^(x-4)

3) Find the value of these logs
a) logv1/2^(1/8)
b) Logv3^(1/3)

4) Which _expression(s) are = to 2?
a) logv3^16 - logv3^4
b) logv4^4^2
c) logv6^1 + logv6^1

5) Use the log of a quotient property to rewrite logv5^(1/8)

6) If y = 8^x, what is its inverse function? My guess is -8^x but who knows. Choices -

a) y = logv8^x
b) y = 8^-x
c) y = -8^x
d) y = 1/8^x

Answer 1

As this seems like homework, I would like to point you on your way before providing you with the answers (since I know if I don't, someone else will anyways ;-)) In that spirit, I have included the answers in the very bottom of the page.

First, number 1. You must first simplify the 3log(4)(x). Use the rule that nlog(a)(b) = log(a)(b^n). Then, you must simplify the subtraction. There is another logarithm rule for that, namely: log(a)(b) - log(a)(c) = log(a)(b/c)

Now, number 2. What you have as the argument to your log function is 5(x-4). Try to expand this using the rule that log(a)(b) + log(a)(c) = log(ac), only reversed and see what you get!.

Number 3. Just remember the definition of a log. That if we have x = log(a)(b), then a^x = b. All of these answers are integers (though they can be negative!) so a little trial and error is all it should take.

Number 4. Some of these do not look easy to do, but first condense them as much as possible. You want a single logarithm! Then, remember that you don't need the exact answer of each logarithm, only if it is equal to 2 or not!

Number 5. The log of a quotient property is just the fact that log(a)(b/c) = log(a)(b) - log(a)(c) as I mentioned earlier on for another problem.

Number 6. You see that the x is an exponent. Then the inverse is a log. The simple way to solve for inverse is to switch the y and x. Now you have x = 8^y. You must solve for y. Look at the definition of a log and see how you can do that.




Answers:

1. Following the rule, it becomes:

3log(4)(x) - log(4)(y) = log(4)(x^3) - log(4)(y) = log(4)[(x^3) / y]

2. Clearly, expanding it, we get:

log(8)[5(x-4)] = log(8)(5) + log(8)(x-4) = (d)

3a. Of course, writing this in exponent form, we have (1/2)^x = (1/8). The answer through trial and error is 3.

3b. This is 3^x = 1/3. The answer would be -1 (you know it is negative since the answer is a fraction and the base is not.

4a. This reduces to:

log(3)(16) - log(3)(4) = log(3)(16 / 4) = log(3)(4) which is not equal to 2 since log(3)(9) = 2.

4b. This reduces to:

log(4)(4^2) = 2log(4)(4) = 2 or
log(4)(4^2) = log(4)(16) = 2.

Either way works.

4c. log(6)(1) + log(6)(1) = 2log(6)(1) also = log(6)(1*1) = log(6)(1). The other way to look at it is to recognize that log(x)(1) where x != 0 is equal to 0! So we have 0 + 0 = 0 != 2.

5. Using it, we get:

log(5)(1) - log(5)(8)

6. We have x = 8^y after switching variables. Accordingly, we solve for y using logs:

y = log(8)(x).

It is difficult to explain as it is the definition. If you look at it, you will see that the exponention form of that log is indeed 8^y = x.


Edit:

@///M5,

Yeah. I actually took a look at the ToS and other parts of the site and could not find anything in there for not helping. I'm quite sure that even if I had not posted, someone would have with just the answers, so I tried to at least point the questioner in the right direction at first, although I may have given away too much. Most of these questions were simple one-step problems so there was not much to give away, lol.

Answer 2

1) Rewrite as a single logarithm: 3 logv4^x - logv4^y

3log(4)x - log(4)y
log(4)(x^3) - log(4)y
log(4)((x^3)/y)

---------------------------------------

2)
log(8)(5(x - 4))
log(8)5 + log(8)(x - 4)

ANS :
d) log(8)5 + log(8)(x - 4)

-----------------------------------------

3) Find the value of these logs
a) (log(1/8))/(log(1/2)) = 3
b) (log(1/3))/(log(3)) = -1

--------------------------------------------

4) Which _expression(s) are = to 2?
a) log(3)16 - log(3)4 = log(3)(16/4) = log(3)(4)

b) log(4)(4^2) = 2log(4)4 = 2(log4 / log4) = 2(1) = 2

c) log(6)1 + log(6)1 = log(6)(1 * 1) = log(6)1 = 0

ANS : b.) log(4)(4^2)

----------------------------------

5) Use the log of a quotient property to rewrite logv5^(1/8)

log(5)(1/8) = log(5)1 - log(8) = 0 - log(8) = -log8

---------------------------------------------

6) If y = 8^x, what is its inverse function? My guess is -8^x but who knows. Choices -

a) y = logv8^x
b) y = 8^-x
c) y = -8^x
d) y = 1/8^x

y = 8^x
x = 8^y
y = log(8)x

ANS : A.) y = log(8)x

The numbers that i have in () right after log can also be written as log(base n)x

Answer 3

2log[9] (anything) = log[3] (the same thing). This comes from the definition of logarithm and the fact that 3^2 = 9. So multiply both sides of your equation by 2 and get: log[3](6x^2 - 19x + 2) = 2log[3](2-3x) or, using a property of logs: log[3](6x^2 - 19x + 2) = log[3](2-3x)^2, so 6x^2 - 19x + 2 = (2-3x)^2, now solve this equation, and be sure to check solutions because solutions (if any) that make either (2 - 3x) or (6x^2 - 19x + 2) negative have to be thrown out, since you cannot take logs of negative numbers.

Answer 4

Dude --

Ask a question or two...but don't have someone do your entire homework for you --

it's not that hard...

and tedjn, you obviously know your logarithm identities -- but don't do someone else's homework for them...?

Answer 5

1.logv4^x*x*x/y
2.d
3.3
4.b
5.sorry I don't understand the question
6.a
At last, good luck! : )

Answer 6

This is misusing yahoo answers.

Kids... start doing you homework yourself...

Don't outsource!!!