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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.