Write the single logarithm as a sum or difference of logs.log3(x9y2)\log_3\left(\frac{\sqrt{x}}{9y^2}\right)log3(9y2x)

12log⁡3x−2−2log⁡3y\frac12\log_3x-2-2\log_3y21log3x−2−2log3y

Write the single logarithm as a sum or difference of logs. log 3 ( x 9 y 2 ) \log_3\left(\frac{\sqrt{x}}{9y^2}\right) lo g 3 ( 9 y 2 x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

In this problem, we're asked to write a single logarithm as a sum or difference of logs. And here, we're given log base 3 of the square root of x divided by 9 y squared, and we wanna go ahead and expand this out as much as possible. So let's go ahead and get started. Now the first thing I notice here is that these numbers are being divided, so I can go ahead and apply the quotient rule and change that division into the subtraction of 2 different logs. So if I expand this out, this becomes a log base 3 of the square root of x minus log base 3 of 9 y squared. Now that I've separated those 2 logs, I'm gonna go to each of these individual logs and see how else I can expand it. Now this log base 3 of the square root of x, I know that I can change the square root of x into an exponent to allow me to use the power rule. So this will become log base 3 of x to the power of onetwo. Now that I have that exponent, I can go ahead and apply the power rule and pull that exponent to the front of my log. So this becomes 1 half times log base 3 of x. And there is nothing else I can do to expand that log, so we are good. This is how this log is going to stay. Now let's go ahead and move on to our other log and focus on that. So looking over here, I have log base 3 of 9 y squared. But since 9 is just 3 squared and my y squared, I can go ahead and pull that squared out into an exponent. So this becomes a minus log base 3 of 3 y squared. And now that I have that exponent, I can again apply the power rule and pull that exponent to the front of my log. Now in doing that, this becomes 2 times log base 3 of 3 y, and that exponent goes away. Now I might want to stop here, but I can actually do even more to expand this because I see that this 3 and this y are multiplying each other. Now whenever we have things being multiplied, we can go ahead and apply the product rule and change that multiplication into the addition of 2 separate logs. Now this one half log base 3 of x is going to stay the same. We're still not worrying about that. Let's go ahead and separate this log base 3 of 3 y out. Now we want to make sure that we keep this negative 2 attached to it because it hasn't gone anywhere. So this is negative 2. And then separating this out using the product rule, this becomes log base 3 of 3 plus log base 3 of y. Now looking at this, I notice this log base 3 of 3. Since my base is the same as what I'm taking the log of, this really just becomes 1. So this is really minus 2 times 1 plus log base 3 of y. And now that I have gotten it to these multiple log expressions, I have simplified it so much, I can go ahead and distribute this 2 into it in order to get my final answer. So I'm gonna go ahead and pull this one half log base 3 of x down because even though we haven't worked with it for a while, it's still a part of our expression. So this is one half log base 3 of x and then a minus I'm gonna go ahead and distribute that to n. So minus 2 and then minus 2 log base 3 of y. Now this is my fully complete answer, and that is my sum or difference of logs that I have gotten from this original, much more compact log expression. Hopefully that makes sense. Let me know if you have questions.

Write the single logarithm as a sum or difference of logs.log⁡3(x9y2)\$$log_3$$\left(\frac{\sqrt{x}}{$$9y^2$$}\right)log3(9y2x)

12log⁡3x−2−2log⁡3y\frac12\log_3x-2-2\log_3y21log3x−2−2log3y

2log⁡3x−2−log⁡39y2\log_3x-2-\log_39y2log3x−2−log39y

12log⁡3x+2log⁡33y\frac12\log_3x+2\log_33y21log3x+2log33y

12log⁡3x−2log⁡39y\frac12\log_3x-2\log_39y21log3x−2log39y

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Precalculus

6. Exponential and Logarithmic Functions

Properties of Logarithms

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Write the single logarithm as a sum or difference of logs.
$$\log$_3$\left$($\frac{\sqrt{x}$}{9y^2}$\right$)$

$2$\log$_3x-2-$\log$_39y$

$$\frac$12$\log$_3x-2-2$\log$_3y$

$$\frac$12$\log$_3x+2$\log$_33y$

$$\frac$12$\log$_3x-2$\log$_39y$

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Start by applying the logarithm property for division: $ \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N $. This allows us to separate the terms inside the logarithm.

Next, apply the logarithm property for roots: $ \log_b \sqrt{M} = \frac{1}{2} \log_b M $. This helps to simplify the square root term inside the logarithm.

Now, apply the logarithm property for powers: $ \log_b M^n = n \log_b M $. This will help to simplify the $ y^2 $ term inside the logarithm.

Combine the results from the previous steps to express the original logarithm as a sum or difference of simpler logarithms.

Ensure all terms are simplified and correctly expressed as a sum or difference of logarithms, using the properties applied in the previous steps.

3:49m

다음 시청

마스터 Product, Quotient, and Power Rules of Logs가 Callie Rethman 짧은 영상으로 설명해 줍니다

Properties of Logarithms 연습 세트

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전문가 영상 설명

Precalculus 때문에 힘들어하고 계신가요?

Write the single logarithm as a sum or difference of logs.log3(x9y2)\(\log\)_3\(\left\)(\(\frac{\sqrt{x}\)}{9y^2}\(\right\))log3(9y2x)

다음 시청

Properties of Logarithms 연습 세트

Write the single logarithm as a sum or difference of logs.
$\(\log\)_3\(\left\)(\(\frac{\sqrt{x}\)}{9y^2}\(\right\))$