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Taya2010 [7]
3 years ago
9

HELP ME NOW I SPENT 99 POINTS TRYING LAST TIEM

Mathematics
1 answer:
marshall27 [118]3 years ago
5 0
2.30 per gallon yghehhf

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If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
Brainliest if you are correct!! Find the domain of the graphed function.
hichkok12 [17]

Answer:

A is correct

Step-by-step explanation:

x must be greater or equal to 0

5 0
3 years ago
Read 2 more answers
3(x-6) -8x= -2+5(2x+1)
topjm [15]

Answer: x= - 7/5

Step-by-step explanation:

did the problem

5 0
2 years ago
Read 2 more answers
Find the rational roots f(x) =3x3+ 2x2 + 3x + 6
Ann [662]

The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

<h3>How to determine the rational root of the function f(x)?</h3>

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

For a function P(x) such that

P(x) = ax^n +...... + b

The rational roots of the function p(x) are

Rational roots = ± Possible factors of b/Possible factors of a

In the function f(x), we have:

a = 3

b = 6

The factors of 3 and 6 are

a = 1 and 3

b = 1, 2, 3 and 6

So, we have:

Rational roots = ±(1, 2, 3, 6)/(1, 3)

Split the expression

Rational roots = ±(1, 2, 3, 6)/1 and ±(1, 2, 3, 6)/3

Evaluate the quotient

Rational roots = ±(1, 2, 3, 6, 1/3, 2/3, 1, 2)

Remove the repetition

Rational roots = ±(1, 2, 3, 6, 1/3, 2/3)

Hence, the rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

The complete parameters are:

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

Read more about rational roots at

brainly.com/question/17754398

#SPJ1

4 0
2 years ago
Thirty students are eating lunch at 5 tables. Each table has the same number of students. How many students are sitting at each
kaheart [24]
30 ÷ 5
6 students at each table

Hope this helps!
5 0
3 years ago
Read 2 more answers
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