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3 years ago

How many triples of three positive integers have a product of 16?

Mathematics

1 answer:

gregori [183]3 years ago

7 0

There are 15 possible triples of three positive integers whose product is 16.

According to this question, we must find the quantity of all <em>possible</em> triples such that they have a product of 16. By factor decomposition, we have that 16 is equal to the following product of <em>prime</em> numbers:

$16 = 2^{4}$

There are the following triples considering order of factors:

(i) $\{16, 1, 1\}$ , (ii) $\{1, 16, 1\}$ , (iii) $\{1, 1, 16\}$ , (iv) $\{2, 8, 1\}$ , (v) $\{8, 2, 1\}$ , (vi) $\{8, 1, 2\}$ , (vii) $\{2, 1, 8\}$ , (viii) $\{1, 2, 8\}$ , (ix) $\{1, 8, 2\}$ , (x) $\{1,4,4\}$ , (xi) $\{4, 1, 4\}$ , (xii) $\{4,4,1\}$ , (xiii) $\{2,2,4\}$ , (xiv) $\{2, 4, 2\}$ , (xv) $\{4,2,2\}$

There are 15 possible triples of three positive integers whose product is 16.

We kindly invite to check this question on factor decomposition: brainly.com/question/2250220

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Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

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In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

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Do this on your calculator to find that

x = 12.99294297

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$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

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x = 5.29449005

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$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

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Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

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But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

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For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

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