Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

For example, lets let k = ln(x)

Answer 6

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

Answer 7

(k)^4= k^4

Answer 8

not quite.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

k^4= 4ln(x) ?

Answer 14

Yes

Answer 15

And since we said k = ln(x) ?

Answer 16

k^4 = 4k?

Answer 17

indeed

Answer 18

So solve for k

Answer 19

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

Answer 20

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

Answer 21

then factor it

Answer 22

Actually that might not work out well

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

That'll work ok

Answer 27

So now lets go back to our definition of k

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

x= (cuberoot 4) e = 4.31?

Answer 32

Should be:

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

Answer 33

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

Answer 34

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

Answer 35

Thanks

Answer 36

Indeed. Takes a bit of thinking