anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
So if you just want a second opinion on your answer, what answer did you get?
anonymous
  • 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
  • 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
  • anonymous
Personally I would start with a substitution to make it easier to see what's going on
anonymous
  • anonymous
For example, lets let k = ln(x)
anonymous
  • anonymous
What would we have for our new version of the equation?
anonymous
  • anonymous
(k)^4= k^4
anonymous
  • anonymous
not quite.
anonymous
  • anonymous
Thats basically the same thing on both sides of the equal sign
anonymous
  • anonymous
It should be \( k^4 = ln(x^4)\)
anonymous
  • anonymous
because in general for some value a, \[ln(a^4) \ne (ln\ a)^4\]
anonymous
  • anonymous
But there is something we can do with the x^4 because of the properties of logarithms
anonymous
  • anonymous
k^4= 4ln(x) ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
And since we said k = ln(x) ?
anonymous
  • anonymous
k^4 = 4k?
anonymous
  • anonymous
indeed
anonymous
  • anonymous
So solve for k
anonymous
  • anonymous
do I divide by 4 first or do I take the 4th root?
anonymous
  • anonymous
I would move it over to the same side so it equals 0.
anonymous
  • anonymous
then factor it
anonymous
  • anonymous
Actually that might not work out well
anonymous
  • anonymous
I mean it's correct, but there might be something easier
anonymous
  • anonymous
hmmmm im trying to factor it and im at k(k^3-4)
anonymous
  • anonymous
Right, so either k = 0 or k^3 = 4
anonymous
  • anonymous
That'll work ok
anonymous
  • anonymous
So now lets go back to our definition of k
anonymous
  • anonymous
ln(x) = 0 and ln(x)^3=4
anonymous
  • anonymous
ln(x) = 0 & ln(x) = (cube root)4 ?
anonymous
  • anonymous
Yep, now raise e to the power of both sides to get rid of the ln
anonymous
  • anonymous
x= (cuberoot 4) e = 4.31?
anonymous
  • anonymous
Should be: \[x = e^0 \text{ or } x = e^\sqrt[3]{4}\]
anonymous
  • anonymous
Because we had \[ln(x) = 0 \text{ or } ln(x) = \sqrt[3]{4}\] and \[ln(a) = b \iff e^b = a\]
anonymous
  • anonymous
5.891 is the answer. That was a tuffy for me.
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
Indeed. Takes a bit of thinking

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