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timofeeve [1]
3 years ago
11

¼ log b x – 2 log b 5 – 10log b y

Mathematics
1 answer:
melamori03 [73]3 years ago
7 0

I will leave out the base, b since it's the same for all these logarithms.

\frac{1}{4} Log(x)-2Log(5)-10Log(y)  \\  \\ Power-Rule \\  \\ Log( x^{ \frac{1}{4} } )-Log( 5^{2} )-Log( y^{10} ) \\  \\ Quotient-Rule \\  \\ Log( x^{ \frac{1}{4} } ) -Log( \frac{25}{ y^{10} } ) \\  \\ Quotient-Rule \\  \\ Log( \frac{ x^{ \frac{1}{4} }  y^{10} }{25} )


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Sketch the graph of each rational function showing all the key features. Verify your graph by graphing the function on
kenny6666 [7]

Answer:

The Answer is:

<em>Domain of the function:</em> Dom= {x ∈ R: x ≠ \frac{-5}{2}}

<em>Horizontal asymptote:</em> y=2

<em>Vertical asymptote:</em> x=\frac{-5}{2}

<em>Cut with X-axis:</em> x=\frac{-6}{5}

Step-by-step explanation:

<em>1. Domain of the function: </em>To find the domain of the function you have to find where the dominator of the function is ZERO, so you have to make 2x+5=0

2x+5=0

2x=-5

x=-5/2 Thats the point of the graph that does NOT exist

The domain of the function is: all real numbers except (-5/2) <em>Dom= {x ∈ R: x ≠ \frac{-5}{2}}</em>

<em>2. Horizontal asymptote: </em> take the first numbers that are with the X's in this case:

4x− 6/ 2x+5 you have to take 4x and 2x so <em>y=4/2</em>

<em></em>

<em>3. Vertical asymptote: </em> take the number of 1. and thats the vertical asymptote in this case is x=-5/2

<em></em>

<em>4. Cut with X-axis: </em> replace the x by zero, f(0) = 4(0) − 6 / 2(0) + 5

f(0)=-6/5,<em> f(x)=-6/5</em>

<em></em>

<em>this are the key features of the graph now you can replace numbers and draw your graph</em>

3 0
4 years ago
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