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

1 2 log ⁡ 3 x − 2 − 2 log ⁡ 3 y \frac12\log_3x-2-2\log_3y 2 1 lo g 3 x − 2 − 2 lo g 3 y

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="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 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 ( 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"> )

1 2 log ⁡ 3 x − 2 − 2 log ⁡ 3 y \frac12\log_3x-2-2\log_3y 2 1 lo g 3 x − 2 − 2 lo g 3 y

2 log ⁡ 3 x − 2 − log ⁡ 3 9 y 2\log_3x-2-\log_39y 2 lo g 3 x − 2 − lo g 3 9 y

1 2 log ⁡ 3 x + 2 log ⁡ 3 3 y \frac12\log_3x+2\log_33y 2 1 lo g 3 x + 2 lo g 3 3 y

1 2 log ⁡ 3 x − 2 log ⁡ 3 9 y \frac12\log_3x-2\log_39y 2 1 lo g 3 x − 2 lo g 3 9 y

6. Exponential & Logarithmic Functions

Properties of Logarithms

College Algebra

6. Exponential & Logarithmic Functions

Properties of Logarithms

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Multiple Choice

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

$2\log_3x-2-\log_39y$

$\frac12\log_3x-2-2\log_3y$

$\frac12\log_3x+2\log_33y$

$\frac12\log_3x-2\log_39y$

Verified step by step guidance

Start by recognizing the expression inside the logarithm: $ \log_3 \left( \frac{\sqrt{x}}{9y^2} \right) $. This is a logarithm of a quotient, which can be expressed as the difference of two logarithms.

Apply the quotient rule for logarithms: $ \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N $. Here, $ M = \sqrt{x} $ and $ N = 9y^2 $.

Rewrite the expression using the quotient rule: $ \log_3 \sqrt{x} - \log_3 (9y^2) $.

Next, apply the power rule for logarithms: $ \log_b (M^n) = n \cdot \log_b M $. For $ \sqrt{x} = x^{1/2} $, this becomes $ \frac{1}{2} \log_3 x $.

Similarly, apply the power rule to $ 9y^2 = 9 \cdot y^2 $, which can be split into $ \log_3 9 + 2 \log_3 y $. Combine these to get the final expression: $ \frac{1}{2} \log_3 x - (\log_3 9 + 2 \log_3 y) $.

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