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Mila [183]
3 years ago
9

URGENT HELP ME PLEASE

Mathematics
1 answer:
Trava [24]3 years ago
5 0

Answer:

(a)\log_3(\dfrac{81}{3})=3

(b)\log_5(\dfrac{625}{25})=2

(c)\log_2(\dfrac{64}{8})=3

(d)\log_4(\dfrac{64}{16})=1

(e)\log_6(36^4)=8

(f)\log(100^3)=6

Step-by-step explanation:

Let as consider the given equations are \log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?.

(a)

\log_3(\dfrac{81}{3})=\log_3(27)

\log_3(\dfrac{81}{3})=\log_3(3^3)

\log_3(\dfrac{81}{3})=3        [\because \log_aa^x=x]

(b)

\log_5(\dfrac{625}{25})=\log_5(25)

\log_5(\dfrac{625}{25})=\log_5(5^2)

\log_5(\dfrac{625}{25})=2        [\because \log_aa^x=x]

(c)

\log_2(\dfrac{64}{8})=\log_2(8)

\log_2(\dfrac{64}{8})=\log_2(2^3)

\log_2(\dfrac{64}{8})=3        [\because \log_aa^x=x]

(d)

\log_4(\dfrac{64}{16})=\log_4(4)

\log_4(\dfrac{64}{16})=1        [\because \log_aa^x=x]

(e)

\log_6(36^4)=\log_6((6^2)^4)

\log_6(36^4)=\log_6(6^8)

\log_6(36^4)=8            [\because \log_aa^x=x]

(f)

\log(100^3)=\log((10^2)^3)

\log(100^3)=\log(10^6)

\log(100^3)=6            [\because \log10^x=x]

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Step-by-step explanation:

The domain of a graph is the span of x-values covered by the graph, while the range of a graph is the span of y-values covered by the graph.

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From the graph, we can see that the x-values covers everything to the left of x=1. Then we stop and then continue beginning at x=2 all the way until positive infinity.

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In interval notation, this is:

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We use brackets with the 1 because we include it in our domain and we use parentheses with the 2 since we do not.

RANGE:

Looking at the graph, it seems that all y-values are covered. The part of the graph on the left covers all y-values less than or equal to 3, while the part of the graph on the right covers all y-values greater than -1. Thus, all y-values are covered.

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In the multiple choice options, all of the answers are in factored form. Factored from for quadratic equations are used to find the zeroes, when you equate each factor to 0.

Remember quadratic equations are in the form ax² + bx + c = 0.

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I will use the <u>cross factors or "X"</u> method.

Write out the factors for "a" and "c" vertically (see diagram 1).

Test a pair of factors from "a" and a pair of factors from "c". Cross multiply them and add their products. If the sum of their products is "b", you found the factors.

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Example 1: (diagram 2)

I will try if the highlighted pairs of factors work. Multiply the pairs diagonally, which gives me "1" and "-60". The sum of "1" and "-60" is -59.

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Example 2: (diagram 3)

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Example 3: (diagram 4)

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The factors are found by reading across the top and the bottom. Include "x" with the first factor.

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