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GenaCL600 [577]
3 years ago
12

Solve log8 12=x-2. Round your answer to four decimal places.

Mathematics
1 answer:
olganol [36]3 years ago
4 0

Answer:

\large\boxed{x\approx3.1950}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\\log_812=x-2\iff8^{x-2}=12\\\\(2^3)^{x-2}=4\cdot3\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{3(x-2)}=2^2\cdot3\qquad\text{use the distributive property}\\\\2^{3x-6}=2^2\cdot3\qquad\text{divide both sides by}\ 2^2\\\\\dfrac{2^{3x-6}}{2^2}=3\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\2^{3x-6-2}=3\\\\2^{3x-8}=3\qquad\text{logarithm of both sides}\\\\\log_22^{3x-8}=\log_23\qquad\text{use}\ \log_aa^n=n\\\\3x-8=\log_23\qquad\text{add 8 to both sides}

3x=\log_23+8\qquad\text{divide both sides by 3}\\\\x=\dfrac{\log_23+8}{3}\\\\\log_23\approx1.5849625\\\\x\approx\dfrac{1.5849625+8}{3}\approx3.1950

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Answer:

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Step-by-step explanation:

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3 years ago
Look at the image. (calculus)
mash [69]
<h3>Answer: Choice H)  2</h3>

=============================================

Explanation:

Recall that the pythagorean trig identity is \sin^2 x + \cos^2x = 1

If we were to isolate sine, then,

\sin^2 x + \cos^2x = 1\\\\\sin^2 x = 1-\cos^2x\\\\\sin x = \sqrt{1-\cos^2x}\\\\

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, \cos x = \sqrt{1-\sin^2x}\\\\

Cosine is also positive in this quadrant.

-------------

So,

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5 0
2 years ago
Read 2 more answers
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