Expand: log8((4√x)/(64y3))

Question

Accepted Answer

Everyone in this problem we are asked to expand using properties of logarithms and the expression were given is log of the square root of a divided by B to the six. Okay so rewriting this log Of the squared of a, divided by 100 B to the six. All right now we have a root in the top, we have an exponent in the bottom. We have some multiplication in the bottom. But the main thing we need to deal with first is the division. Okay, so how do we deal with division? Let's recall our log property for dealing with division. If we have log of base B of some quantity X divided by some quantity Y. We can write this as log base B of X minus log base via boy. Okay so the division within the log becomes the subtraction of two logs applying this to our expression. Okay we're going to get log of the square root of a. Okay, the numerator minus log of the denominator which in this case is 100 B to the six. Alright, great. Now looking at this first term here, we have a route. We know we have a log exponent rule. A rule that will allow us to do with exponents and logs and we also know that we have a rule that allows us to write roots in terms of exponents. Okay, so let's first write this route in terms of an exponent case. We're going to use this rule here if we have the root of M. We can write this as M to the one over end. Okay so we're gonna apply that to the first term here. Now looking at this second term while here we have multiplication. Okay We have 100 times B to the six within a log. Now let's recall our log roll for dealing with multiplication. If we have log base B. Of X times Y. We can write this as log base B. Of X plus log base B. Of why? Alright, so the base stays the same and the multiplication within the log becomes the addition of two logs. Alright so applying both of these rules, the first term, we're going to have log of A to the one half. And in the second term we're gonna subtract, we're going to put a big square bracket here. Okay because this entire log 100 B to the six term is being subtracted. So when we separate this using the next rule we need to make sure that that negative applies to both terms. Okay, so big square bracket here so that we don't mix up our signs and we're gonna get log of 100 plus log B to the six. Alright great. Now let's go ahead and expand and just distribute that negative. We're also going to Right we have 10 or sorry 100 here. The base of this log again. If it's not written, recall that the base is 10. Okay, so the base is 10 here. Let's try to write 100 with a base of 10. We know we can do that. 10 squared is 100. So let's do that as well. So we have log of eight of the one half Minour log of 10 squared plus scoops and minus. And this is exactly why we put that big bracket. So we didn't mix up that sign minus log of B to the six. All right now we have an exponent here in every single term. So let's go ahead and deal with that. If we have log base B, that's some X to the exponent end, recall that we can write this as n log base B of X to the exponent end can come out in front and be multiplied by the log. Alright so applying this in the first term we're gonna have 1/ log of a -2. long of 10 - Log B. Alright, now, this first term there's nothing more. We can expand here. Same with the last term the minus six log B. However the minus two log 10, remember that If we don't write the base it's assumed to be 10. Okay? So if we have log base 10 of 10. Well that's just equal to one. So we can write this as one half log of a -2 -6 Log B. Okay, there's nothing more. We can expand and so this is going to be our answer here. One half log a minus two minus six log B. And that's going to correspond with answer A. I hope this video helped. Thanks everyone for watching.

Struggling with College Algebra?

Expand: log8((4√x)/(64y3))

Watch next

Properties of Logarithms practice set