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anonymous
 5 years ago
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
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if you just want a second opinion on your answer, what answer did you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you just want to get the answer, use wolframalpha. If you want to learn you'll need to put in some effort.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually I have no idea how to solve this one. Im not sure what they want. Any ideas?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Personally I would start with a substitution to make it easier to see what's going on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For example, lets let k = ln(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What would we have for our new version of the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thats basically the same thing on both sides of the equal sign

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should be \( k^4 = ln(x^4)\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because in general for some value a, \[ln(a^4) \ne (ln\ a)^4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But there is something we can do with the x^4 because of the properties of logarithms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And since we said k = ln(x) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do I divide by 4 first or do I take the 4th root?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would move it over to the same side so it equals 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually that might not work out well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean it's correct, but there might be something easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmm im trying to factor it and im at k(k^34)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, so either k = 0 or k^3 = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now lets go back to our definition of k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(x) = 0 and ln(x)^3=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(x) = 0 & ln(x) = (cube root)4 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep, now raise e to the power of both sides to get rid of the ln

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x= (cuberoot 4) e = 4.31?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should be: \[x = e^0 \text{ or } x = e^\sqrt[3]{4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because we had \[ln(x) = 0 \text{ or } ln(x) = \sqrt[3]{4}\] and \[ln(a) = b \iff e^b = a\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05.891 is the answer. That was a tuffy for me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed. Takes a bit of thinking
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