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2 years ago

Write the expression as a single logarithm

Mathematics

1 answer:

vredina [299]2 years ago

4 0

$3\log_7y + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - \log_7z^{5})\\\log_7y^{3} + 4(\log_7(\frac{x}{z^{5}}))\\\log_7y^{3} + \log_7(\frac{x}{z^{5}})^{4}\\\log_7(y^{3} \times (\frac{x}{z^5{5}})^{4})$

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mash [69]

Answer:

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Step-by-step explanation:

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2 years ago

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2 years ago

Can someone answer this plz (-2z^2-14)-(-12z^2)

Orlov [11]

Answer:

$\large\boxed{(-2z^2-14)-(-12z^2)=10z^2-14}$

Step-by-step explanation:

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3 years ago

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maria [59]

Step-by-step explanation:

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2 years ago

Read 2 more answers

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

6 0

2 years ago