Rewrite the log expression as the sum or difference of multiple logs.log2(3a2b45c4)\log_2\left(\frac{3a^2b^4}{\sqrt{5c^4}}\right)

log⁡23+2log⁡2a+4log⁡2b−12log⁡25−2log⁡2c\$$log_23+2$$\log_2a+4\log_2b-\frac12\$$log_25-2$$\log_2clog23+2log2a+4log2b−21log25−2log2c

Rewrite the log expression as the sum or difference of multiple logs. log 2 ( 3 a 2 b 4 5 c 4 ) \log_2\left(\frac{3a^2b^4}{\sqrt{5c^4}}\right)

Here we're asked to rewrite the log expression as the sum or difference of multiple logs. Here we have log base two of three a squared b to the fourth over the square root of five c to the fourth. Now we want to expand this as much as possible, and there is quite a bit going on here, but let's break this down step by step. Looking at this argument, the first thing that I see is that I have division happening. So I have three a squared b to the fourth being divided by the square root of five c to the fourth. So I'm gonna start here by applying the quotient property. So rewriting this as a log base two of three a squared b to the fourth minus log base two of the square root of five c to the fourth. So now we've broken this down and we have two slightly more simple logs, and we can focus on each of those individually to get our final answer. I'm gonna start here by focusing on this first log expression, log base two of three a squared b to the fourth. So here in the argument of this log, I have three multiplying a squared, multiplying b to the fourth power. So this gives me a log base two of three plus log base two of a squared plus log base two of b to the fourth. So breaking that out using the product property. Now I can apply the power property as well, pulling those exponents, those powers out to the front, multiplying by them. So this gives me a log base two of three plus two times log base two of a plus four times log base two of b. So I've taken that first log and I've expanded it out into three separate logs And now I need to focus on this second log here. So this log base two of the square root of five C to the fourth power. Now, before I do anything with this, I need to take care of that square root. Remember that a square root is the same thing as taking something and raising it to the power of one half. So I want to go ahead and apply my power property here before I can do any further expansion. So I'm gonna pull that one half out front, multiplying by it. So I now have minus one half log base two of five c to the fourth power. Now I can apply the product property again here now for this log because five is multiplying c to the fourth power. Now I wanna make sure to take account of this negative one half out front. So I have minus one half and now that I'm expanding this log further, I'm gonna put a bracket there just to make sure that I know that this negative one half applies to everything that I'm gonna write here. So I have log base two of five plus log base two of c to the fourth power. So I expanded this out using the product property. I have one final thing I can do in expanding this and that's pull that four out front, multiplying by it using the power property. So I now have a minus one half log base two of five plus four times log base two of c. And now that I've done everything I can with my log properties, I just need to do some algebra in order to further simplify this, going ahead and taking this negative one half and distributing it to each of these terms. So that gives me negative one half log base two of five and then negative one half multiplying it by four log base two of c gives me minus two log base two of c. So that's now this second log expanded out into these two logs. Now we want to put this all together to get our final answer here, so I'm going to go ahead and copy this and put it together with that final answer. And this is my fully expanded log expression. So I started with log base two of three a squared b to the fourth over the square root of five c to the fourth. And I ended up with log base two of three plus two log base two of a plus four log base two of b minus one half log base two of five minus two log base two of c. So that is expanded as much as possible. Let us know if you have any questions, and I'll see you in the next one.

Rewrite the log expression as the sum or difference of multiple logs.log⁡2(3a2b45c4)\$$log_2$$\left(\frac{$$3a^2$b^4$$}{\sqrt{$$5c^4$$}}\right)

log⁡23+2log⁡2a+4log⁡2b−12log⁡25−2log⁡2c\$$log_23+2$$\log_2a+4\log_2b-\frac12\$$log_25-2$$\log_2clog23+2log2a+4log2b−21log25−2log2c

11. Inverse, Exponential, & Logarithmic Functions

Properties of Logarithms

Intermediate Algebra

11. Inverse, Exponential, & Logarithmic Functions

Properties of Logarithms

Difficoltà con Intermediate Algebra?

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Rewrite the log expression as the sum or difference of multiple logs.
$\log_{2} (\frac{3 a^{2} b^{4}}{\sqrt{5 c^{4}}})$

$$\log$_23+2$\log$_2a+4$\log$_2b-$\frac$12$\log$_25-2$\log$_2c$

$$\log$_23+2$\log$_2a+4$\log$_2b-$\frac$12$\log$_25+4$\log$_2c$

$$\log$_23+2$\log$_2a+4$\log$_2b-2$\log$_25-$\frac$12$\log$_2c$

$$\log$_23+2$\log$_2a+4$\log$_2b-2$\log$_25+$\frac$12$\log$_2c$

Guida verificata passo dopo passo

Start with the given logarithmic expression: $\log_2\left(\frac{3a^2b^4}{\sqrt{5c^4}}\right)$.

Use the logarithm quotient rule to rewrite the log of a fraction as the difference of two logs: $\log_2(\text{numerator}) - \log_2(\text{denominator})$. So, write it as $\log_2(3a^2b^4) - \log_2(\sqrt{5c^4})$.

Apply the logarithm product rule to each log of a product, expressing it as a sum of logs. For the numerator: $\log_2 3 + \log_2 a^2 + \log_2 b^4$. For the denominator inside the square root, first rewrite the square root as an exponent: $\sqrt{5c^4} = (5c^4)^{1/2}$.

Use the power rule of logarithms to bring exponents in front of the logs. For the numerator terms: $\log_2 a^2 = 2 \log_2 a$ and $\log_2 b^4 = 4 \log_2 b$. For the denominator: $\log_2 (5c^4)^{1/2} = \frac{1}{2} \log_2 (5c^4)$, which further breaks down into $\frac{1}{2} (\log_2 5 + \log_2 c^4)$.

Finally, apply the power rule again to $\log_2 c^4 = 4 \log_2 c$, and combine all terms carefully, remembering to subtract the entire denominator expression from the numerator expression.