Solve for x exactly (no graphing) in (ln x)^4 = ln x^4. You will get two answers. Their sum is approximately
A) 4.291
B) 4.691
C) 5.091
D) 5.491
E) 5.891

- anonymous

- schrodinger

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- anonymous

So if you just want a second opinion on your answer, what answer did you get?

- anonymous

If you just want to get the answer, use wolframalpha. If you want to learn you'll need to put in some effort.

- anonymous

Actually I have no idea how to solve this one. Im not sure what they want. Any ideas?

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## More answers

- anonymous

Personally I would start with a substitution to make it easier to see what's going on

- anonymous

For example, lets let k = ln(x)

- anonymous

What would we have for our new version of the equation?

- anonymous

(k)^4= k^4

- anonymous

not quite.

- anonymous

Thats basically the same thing on both sides of the equal sign

- anonymous

It should be
\( k^4 = ln(x^4)\)

- anonymous

because in general for some value a,
\[ln(a^4) \ne (ln\ a)^4\]

- anonymous

But there is something we can do with the x^4 because of the properties of logarithms

- anonymous

k^4= 4ln(x) ?

- anonymous

Yes

- anonymous

And since we said k = ln(x) ?

- anonymous

k^4 = 4k?

- anonymous

indeed

- anonymous

So solve for k

- anonymous

do I divide by 4 first or do I take the 4th root?

- anonymous

I would move it over to the same side so it equals 0.

- anonymous

then factor it

- anonymous

Actually that might not work out well

- anonymous

I mean it's correct, but there might be something easier

- anonymous

hmmmm im trying to factor it and im at k(k^3-4)

- anonymous

Right, so either k = 0 or k^3 = 4

- anonymous

That'll work ok

- anonymous

So now lets go back to our definition of k

- anonymous

ln(x) = 0 and ln(x)^3=4

- anonymous

ln(x) = 0 & ln(x) = (cube root)4 ?

- anonymous

Yep, now raise e to the power of both sides to get rid of the ln

- anonymous

x= (cuberoot 4) e = 4.31?

- anonymous

Should be:
\[x = e^0 \text{ or } x = e^\sqrt[3]{4}\]

- anonymous

Because we had
\[ln(x) = 0 \text{ or } ln(x) = \sqrt[3]{4}\]
and
\[ln(a) = b \iff e^b = a\]

- anonymous

5.891 is the answer. That was a tuffy for me.

- anonymous

Thanks

- anonymous

Indeed. Takes a bit of thinking

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