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3 years ago

Can someone please write this as a single logarithm and show work please and thank you

Mathematics

1 answer:

Mazyrski [523]3 years ago

4 0

Answer:

$log_b(w^3y^4)$

Step-by-step explanation:

To write the expression as a single logarithm, or condense it, use the properties of logarithms.

1) The power property of logarithms states that $log_ax^r = rlog_ax$ . In other words, the exponent within a logarithm can be brought out in front so it's multiplied by the logarithm. This means that the number in front of the logarithm can also be brought inside the logarithm as an exponent.

So, in this case, we can move the 3 and the 4 inside the logarithms as exponents. Apply this property as seen below:

$3log_bw+4log_by\\=log_bw^3+log_by^4$

2) The product property of logarithms states that $log_axy = log_ax+log_ay$ . In other words, the logarithm of a product is equal to the sum of the logarithms of its factors. So, in this case, write the expression as a single logarithm by taking the log (keep the same base) of the product of $w^3$ and $y^4$ . Apply the property as seen below and find the final answer.

$log_bw^3+log_by^4\\=log_b(w^3y^4)$

So, the answer is $log_b(w^3y^4)$ .

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A survey showed that 82% of youth most often used the internet at home. What fraction of youth surveyed most often use the inter

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Step-by-step explanation:

At first, this problem might seem confusing since there could be places other than home with different percentages, and it feels as if you don't know enough to solve the problem.

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3 years ago

write an equation for an ellipse centered at the origin, which has foci at (+-3,0) and co vertices at (0+-4)

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Answer:

The equation for an ellipse centered at the origin with foci at (-3, 0) and (+3, 0) and co-vertices at (0, -4) and (0, +4) is:

$\frac{x^{2}}{7} + \frac{y_{2}}{16} = 1$

Step-by-step explanation:

An ellipse center at origin is modelled after the following expression:

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Where:

$a$ , $b$ - Major and minor semi-axes, dimensionless.

The location of the two co-vertices are (0, - 4) and (0, + 4). The distance of the major semi-axis is found by means of the Pythagorean Theorem:

$2\cdot b = \sqrt{(0-0)^{2}+ [4 - (-4)]^{2}}$

$2\cdot b = \pm 8$

$b = \pm 4$

The length of the major semi-axes can be calculated by knowing the distance between center and any focus (c) and the major semi-axis. First, the distance between center and any focus is determined by means of the Pythagorean Theorem:

$2\cdot c = \sqrt{[3 - (-3)]^{2}+ (0-0)^{2}}$