2log(2a - 1) = 0 How

Mathematics

1 answer:

Kryger [21]2 years ago

3 0

Answer:

Step-by-step explanation:

$2\log(2a-1)=0\\\\\text{Domain:}\\\\2a-1>0\qquad\text{add 1 to both sides}\\2a-1+1>0+1\\2a>1\qquad\text{divide both sides by 2}\\\dfrac{2a}{2}>\dfrac{1}{2}\\\boxed{a>0.5}\\============================\\\\2\log(2a-1)=0\qquad\text{divide both sides by 2}\\\\\dfrac{2\!\!\!\!\diagup\log(2a-1)}{2\!\!\!\!\diagup}=\dfrac{0}{2}\\\\\log(2a-1)=0\qquad\text{use}\ \log_ab=c\iff b=a^c\\\\\log(2a-1)=\log10^0\\\\\log(2a-1)=\log1\iff2a-1=1\qquad\text{add 1 to both sides}\\\\2a-1+1=1+1\\\\2a=2\qquad\text{divide both sides by 2}\\\\\dfrac{2a}{2}=\dfrac{2}{2}\\\\a=1\in D$

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Let A = (1, 0, 3) and B = (−3, 2, 1).

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a) $(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

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c) $-2x+y-z=1$

Step-by-step explanation:

AB, OB, OA are vectors.

$AB = OB - OA = (-3, 2, 1) - (1, 0, 3) = (-4, 2, -2)\\||AB|| = \sqrt{(-4)^2+2^2+(-2)^2}=\sqrt{24}\\$

Normalizing AB: $\frac{AB}{||AB||} = (\frac{-4}{\sqrt{24}},\frac{2}{\sqrt{24}}, \frac{-2}{\sqrt{24}})=(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$