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Mazyrski [523]
3 years ago
5

[20 pts] evaluate the logarithm log(6)1/36​, show work pls!

Mathematics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

\large\boxed{\log_6\dfrac{1}{36}=-2}

Step-by-step explanation:

\text{We know:}\\\\\log_ab=c\iff a^c=b\\\\\log_6\dfrac{1}{36}\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\=\log_636^{-1}=\log_6(6^2)^{-1}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\log_66^{(2)(-1)}=\log_66^{-2}\qquad\text{use}\log_ab^n=n\log_ab\\\\=-2\log_66\qquad\text{use}\ \log_aa=1\\\\=-2(1)=-2

\log_6\dfrac{1}{36}=-2\ \text{because}\ 6^{-2}=\dfrac{1}{6^2}=\dfrac{1}{36}

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4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

Step-by-step explanation:

We know that:

4p\left(y-k\right)=\left(x-h\right)^2 is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

x^2\:+\:10x\:+\:6y\:=\:47

Rewriting the equation in the standard form

4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by x^2\:+\:10x\:+\:6y\:=\:47 is:

4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

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vertex (h, k) = (-5, 12)

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