Write the single logarithm as a sum or difference of logs.log5(5(2x+3)2x3)\log_5\left(\frac{5\left(2x+3\right)^2}{x^3}\right)log5(x35(2x+3)2)

1+2log⁡5(2x+3)−3log⁡5x1+2\$$log_5$$\left(2x+3\right)-3\log_5x1+2log5(2x+3)−3log5x

Write the single logarithm as a sum or difference of logs. log 5 ( 5 ( 2 x + 3 ) 2 x 3 ) \log_5\left(\frac{5\left(2x+3\right)^2}{x^3}\right) lo g 5 ( x 3 5 ( 2 x + 3 ) 2 )

In this problem, we're asked to take the single logarithm and write it as a sum or difference of logs. And here we have a rather complicated single log. We have log base 5 of 5 times 2x plus 3 squared over x cubed. But we're just going to take it step by step in order to expand this fully. So looking at this, I notice a couple of different things. I notice that I have this 5 multiplying this expression, and then I also am being divided by x cubed. So I can apply both the product and the quotient rule here, and you can choose to do these 1 at a time, but I'm gonna go ahead and do them both at the same time. So I know that with the product rule, this is going to become addition of 2 separate logs. And then with the quotient rule, I'm going to have subtraction. So let's go ahead and expand this out. So first dealing with our product rule, this is log base 5 of 5 and then a plus log base 5 of 2x+3 squared. Now dealing with my quotient rule, I'm gonna go ahead and subtract what's on the bottom of my fraction there because it's being divided. So this becomes a minus log base 5 of x cubed. Okay. Now that we have 3 separate logs, let's go into each of these logs and see how else we can expand or simplify this. Looking at this log base 5 of 5, since my base is the same as what I'm taking the log of, this is really just 1. So I don't have to worry about that at all because that cannot be expanded anymore. It's just 1. Let's move on to our next log. We have log base 5 of 2x +3squared. Now here, I can go ahead and pull this exponent to the front of that expression. So this becomes plus 2 times log base 5 of 2x+ 3. Now looking at this, you might be tempted to separate this out into multiple logs. But since this is being added, we can't do anything else to expand this here. So this is done. This is 2 times log base 5 of 2x plus 3. We can't expand that anymore. I know you might be tempted to, but that's it. Remember that our product rule is for if we have multiplication inside of our log and not addition. So I can't do anything else here. Now looking to my last log here, I have log base 5 of x cubed, and I again see that I have an exponent that I can go ahead and pull to the front of my log using the power rule. So this becomes a minus 3 times log base 5 of x. And looking at this, I don't have any other ways that I can expand this. So I'm actually completely done. I have 1 +2logbase5of2xplus3-3logbase5ofx and we are done here. We have taken our single log. And we now have a sum or difference of logs. Thanks for watching. And let me know if you have questions.

Write the single logarithm as a sum or difference of logs.log⁡5(5(2x+3)2x3)\$$log_5$$\left(\frac{5\left(2x+3\$$right)^2$$}{$$x^3$$}\right)log5(x35(2x+3)2)

1+2log⁡5(2x+3)−3log⁡5x1+2\$$log_5$$\left(2x+3\right)-3\log_5x1+2log5(2x+3)−3log5x

5+2log⁡5(2x+3)−log⁡53x5+2\$$log_5$$\left(2x+3\right)-\log_53x5+2log5(2x+3)−log53x

2log⁡5(2x+3)−3log⁡5x2\$$log_5$$\left(2x+3\right)-3\log_5x2log5(2x+3)−3log5x

log⁡5(2x+3)−log⁡5x\$$log_5$$\left(2x+3\right)-\log_5xlog5(2x+3)−log5x

6. Exponential & Logarithmic Functions

Properties of Logarithms

College Algebra

6. Exponential & Logarithmic Functions

Properties of Logarithms

Struggling with College Algebra?

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Multiple Choice

Write the single logarithm as a sum or difference of logs.
$$\log$_5$\left$($\frac{5\left(2x+3\right)^2}{x^3}\]\right$)$

$5+2$\log$_5$\left$(2x+3$\right$)-$\log$_53x$

$2$\log$_5$\left$(2x+3$\right$)-3$\log$_5x$

$1+2$\log$_5$\left$(2x+3$\right$)-3$\log$_5x$

$$\log$_5$\left$(2x+3$\right$)-$\log$_5x$

Verified step by step guidance

Start by applying the properties of logarithms to the given expression: $ \log_5\left(\frac{5(2x+3)^2}{x^3}\right) $. The logarithm of a quotient can be expressed as the difference of two logarithms: $ \log_5(a/b) = \log_5(a) - \log_5(b) $.

Apply the quotient rule: $ \log_5\left(\frac{5(2x+3)^2}{x^3}\right) = \log_5(5(2x+3)^2) - \log_5(x^3) $.

Next, apply the product rule to the first term: $ \log_5(5(2x+3)^2) = \log_5(5) + \log_5((2x+3)^2) $. The product rule states that $ \log_5(ab) = \log_5(a) + \log_5(b) $.

Now, apply the power rule to the term $ \log_5((2x+3)^2) $: $ \log_5((2x+3)^2) = 2\log_5(2x+3) $. The power rule states that $ \log_5(a^b) = b\log_5(a) $.

Finally, apply the power rule to the term $ \log_5(x^3) $: $ \log_5(x^3) = 3\log_5(x) $. Combine all the terms: $ \log_5(5) + 2\log_5(2x+3) - 3\log_5(x) $. Since $ \log_5(5) = 1 $, the expression simplifies to $ 1 + 2\log_5(2x+3) - 3\log_5(x) $.

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Write the single logarithm as a sum or difference of logs.log5(5(2x+3)2x3)\(\log\)_5\(\left\)(\(\frac{5\left(2x+3\right)^2}{x^3}\]\right\))log5(x35(2x+3)2)

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Properties of Logarithms practice set

Write the single logarithm as a sum or difference of logs.
$\(\log\)_5\(\left\)(\(\frac{5\left(2x+3\right)^2}{x^3}\]\right\))$