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3 years ago

Find the Domain of the following exponential and logarithmic equations and solve them:

Mathematics

1 answer:

kakasveta [241]3 years ago

5 0

The base of a logarithm should always be positive and can't be equal to 1, so the domain is 0 < x < 1 or x > 1.

$\log_{\frac1x}243=5$

Write both sides as powers of 1/x :

$\left(\dfrac1x\right)^{\log_{\frac1x}243}=\left(\dfrac1x\right)^5$

Recall that $a^{\log_ab}=b$ , so that

$243=\left(\dfrac1x\right)^5$

$243=\dfrac1{x^5}$

$x^5=\dfrac1{243}$

Take the 5th root of both sides, recalling that 3⁵ = 243, so

$x=\sqrt[5]{\dfrac1{243}}=\boxed{\dfrac13}$

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babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function. Watch and see.

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Take the inverse sin of both sides. When you do that, you are left with just 2theta on the left. That's why you do this.

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This simplifies to

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We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2. Those are:

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$2\theta=\frac{7\pi }{6}$

Divide both sides by 2 in both of those equations to get that values of theta are:

$\theta=\frac{5\pi }{12},\frac{7\pi }{12}$

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Take the inverse tangent of both sides:

$tan^{-1}(tan(7a))=tan^{-1}(1)$

Taking the inverse tangent of the tangent on the left leaves us with just 7a. This simplifies to

$7a=tan^{-1}(1)$

We look to the unit circle to find which values of a give us a tangent of 1. They are:

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Take the inverse cosine of each side. The inverse cosine and cosine undo each other, leaving us with just 3beta on the left, just like in the previous problems. That simplifies to:

$3\beta=cos^{-1}(\frac{1}{2})$

We look to the unit circle to find the values of beta that give us the cosine of 1/2 and those are:

$3\beta =\frac{\pi}{6},3\beta =\frac{5\pi}{6}$

Divide each of those by 3 to find the values of beta are:

$\beta =\frac{\pi }{18} ,\frac{5\pi}{18}$

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Let's rewrite this in terms of a trig ratio that we are a bit more familiar with:

$\frac{1}{cos(3\alpha) } =\frac{-2}{1}$

We are going to simplify this even further by flipping both fraction upside down to make it easier to solve:

$cos(3\alpha)=-\frac{1}{2}$

Now we will take the inverse cos of each side (same as above):

$3\alpha =cos^{-1}(-\frac{1}{2} )$

We look to the unit circle one last time to find the values of alpha that give us a cosine of -1/2:

$3\alpha =\frac{7\pi}{6},3\alpha =\frac{11\pi}{6}$

Dividing both of those equations by 3 gives us

$\alpha =\frac{7\pi}{18},\frac{11\pi}{18}$

And we're done!!!

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