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# Logs watch

1. Hello

Solving for x:

1. e^4x - e^x=0 (I got 0)

2. log(base3)x - 4log(base x)3 + 3 = 0

For the second I brought it down to log(base3)27x = 4/(log(base3)x). Not sure if it's correct up to there.

2. 1. Fine. (Why?)
2. Leave it as log x + 3 = 4/log x. Let t = log x; now you have t + 3 = 4/t. Do you know how to solve this for t? (Try multiplying through by t first.)
3. (Original post by azamally)
Hello

Solving for x:

1. e^4x - e^x=0 (I got 0)

2. log(base3)x - 4log(base x)3 + 3 = 0

For the second I brought it down to log(base3)27x = 4/(log(base3)x). Not sure if it's correct up to there.

By inspection I think it is fairly obvious that
4. (Original post by steve2005)
I think it is fairly obvious
I see. Why is it fairly obvious that there aren't any other solutions?
5. (Original post by generalebriety)
I see. Why is it fairly obvious that there aren't any other solutions?
It is obvious that one of the solutions is I have not said anything about other possible solutions.

Incidentally in your earlier post you have suggested replacing log(base3) x and log(base x) 3 with t. Note they are different bases.
6. (Original post by steve2005)
It is obvious that one of the solutions is I have not said anything about other possible solutions.

Incidentally in your earlier post you have suggested replacing log(base3) x and log(base x) 3 with t. Note they are different bases.
No I haven't: logx(3) = 1/log3(x). All of the logarithms in my first post are to the base 3.
7. (Original post by generalebriety)
No I haven't: logx(3) = 1/log3(x). All of the logarithms in my first post are to the base 3.

I have two values for x

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