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truBlu · Oct 17th, 2007, 12:43 PM

I know this isn't a math board but I've decided to post a question here anyway. Now before you guys start shouting that I want you to do my HW for me, this particular problem has me completely at a loss. In my book, the answers are listed as .01 and 100 but I can't figure out how you would come up with those answers. I tried taking the log of both sides but it results in an answer of 0 = 2 which is clearly incorrectly. Please take a look and see if you can find the solution or at least point me in the right direction.

(Toggle Plain Text)

x^(logx) = 100x

SamReidHughes · Oct 17th, 2007, 1:09 PM

If you take the log of both sides, you get

log (x ^ (log x)) = log (100 * x)
i.e.
log (x) * log (x) = log (100) + log (x)

Now you have a quadratic equation to solve.

truBlu · Oct 17th, 2007, 1:29 PM

Ugh. Power to a power. It all makes sense now. I was taking log(x^(logx)) and coming up with logx. Thank you.

Oct 17th, 2007, 12:43 PM	#1
truBlu Newbie Join Date: Sep 2007 Location: New Jersey Posts: 13 Rep Power: 0	Logarithmic equation I know this isn't a math board but I've decided to post a question here anyway. Now before you guys start shouting that I want you to do my HW for me, this particular problem has me completely at a loss. In my book, the answers are listed as .01 and 100 but I can't figure out how you would come up with those answers. I tried taking the log of both sides but it results in an answer of 0 = 2 which is clearly incorrectly. Please take a look and see if you can find the solution or at least point me in the right direction. (Toggle Plain Text) x^(logx) = 100x x^(logx) = 100x

Oct 17th, 2007, 1:09 PM	#2
SamReidHughes Newbie Join Date: Oct 2007 Posts: 13 Rep Power: 0	Re: Logarithmic equation If you take the log of both sides, you get log (x ^ (log x)) = log (100 * x) i.e. log (x) * log (x) = log (100) + log (x) Now you have a quadratic equation to solve.

Oct 17th, 2007, 1:29 PM	#3
truBlu Newbie Join Date: Sep 2007 Location: New Jersey Posts: 13 Rep Power: 0	Re: Logarithmic equation Ugh. Power to a power. It all makes sense now. I was taking log(x^(logx)) and coming up with logx. Thank you.

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Logarithmic equation