Question

Use the properties of logarithms to rewrite the expressions. Simplify the result if possible?
Assume all variables represent positive real numbers.
6) logb 3√(x6/y7z8 )
a) 6lobbx-(7logby-​
B) 1/3logbx6-logby7-logbz8
C) 2logbx÷(7/3 logby×8/3logbz)​
D) 2logbx-7/3logby-8/3logbz
Please explain the steps!!

Answer 1

God, these look crazy when they aren't written on paper, huh?
Let's start with the first one,
\[\log_{b} \sqrt[3]{\frac{ x6 }{y7 \times x8 }}\]
First thing we have to do, is get rid of that cubed root. We start by distributing it to everywhere in the parentheses, and we end up with \[\log_{b} \frac{ 6x^{\frac{ 1 }{ 3 }} }{ 7y ^{\frac{ 1 }{ 3 } }\times 8z ^{\frac{ 1 }{ 3 }} }\]
Do you get it up to here?

Answer 2

Also, we are basically expanding the equation here, right?

Answer 3

Yes, but can please explain step by step. This problem is driving me nuts.

Answer 4

Well, I'm mainly just trying to get to the part where your having trouble with. Do you understand what I did with the cubed root?

Answer 5

Not really :(

Answer 6

Alright, so this is something that's required to simplify a lot of these. If you have any root, squared, cubed, or whatever in \[\sqrt[n]{x}\]
You can rewrite it as whatever is under the radical to the power of \[\frac{ 1 }{ n }\]
So we can rewrite \[\sqrt[n]{x}\] As \[x^{\frac{ 1 }{ n }}\]
Sorry it's so hard to read, but do you understand that?

Answer 7

Oh, yes! What's next?

Answer 8

Alright, I'm lazy, so I don't want to rewrite that last equation way up there, but now we have to write it out in the form of logb () + ...
You know, in the structure of almost all the other ones. I assume you have already gone over the properties of logs?
Any variable - coefficient pair in the numerator (when there is only a fraction left), turns into a positive log, and variable coefficient pair in the denominator turns into it's own negative log. So, you get this equation :
\[\log_{b} 6x ^{\frac{ 1 }{ 3 }} - \log_{b} 7y^{\frac{ 1 }{ 3 }} - \log_{b} 8z ^{\frac{ 1 }{ 3 }}\]
Do we get up to this far?

Answer 9

Seriously, this one is a pain to get, but I think you can do it.

Answer 10

Okay I get it so far :)!

Answer 11

The next thing we do, is shift the exponents into the front, so we get:
\[\frac{ 1 }{ 3 } \times \log_{b} 6x - \frac{ 1 }{ 3 } \log_{b} 7y - \frac{ 1 }{ 3 } \log_{b} 8z \]
And that's it, the equation is fully expanded.

Answer 12

Do you get that? Because these next ones are condensing, which is basically the opposite.

Answer 13

I did :)

Answer 14

Alright, now I'm gonna work through with you on this first condensing one, then I want you to try the next ones, ok?

Answer 15

Alright :)

Answer 16

Speaking of the next one, A, it seems cut off, can you rewrite it?

Answer 17

A) 6 log bX-7log b Y-8log b Z

Answer 18

One more property I'm gonna put in here so I don't forget... loga x = ( logb x ) / ( logb a )

Answer 19

Alright, so the first thing we do is take the multiplication bits at the beginning and convert them to exponents, so we get \[\log_{b}x ^{6} - \log_{b} y ^{7} - \log_{b} z ^{8} \]
The final thing we have to do is simply put the variables with a positive log on top and the one with negative logs on the bottom. That will make \[\log_{b}\left( \frac{ x ^{6} }{ y ^{7} \times z ^{8} } \right) \] and it's condensed! do you follow that? If you do, I want you to try to get the next one on your own.

Answer 20

Also, for this next one, keep in mind if there is already an exponent there, when we shift the number at the front of the log it is multiplied by the existing exponent.

Answer 21

I'm sorry what do you mean by "next"? I got a bit lost again

Answer 22

I'm sorry, I worded it confusingly. I want you to try part b, and you probably will notice that there is a pre-existing exponent and also a multiplier in the front of the log. You have to multiply the multiplier and the exponent to get the final exponent. If you don't get what I'm saying now, you'll get it when you start part b.