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$${ x }^{ 3-\log { x } }=100$$

I couldn't think up of an algebraic way to solve this equation so I resorted to thinking through it logically. Although this led me to the correct solution, I don't think that this method is concrete enough to use. How can I solve this properly?

Steps I took to solve this my way:

$${ x }^{ 3-\log { x } }=100^{ 1 }\quad or\quad { x }^{ 3-\log { x } }=10^{ 2 }$$

$$3-\log { x } =1\quad or\quad 3-\log { x } =2$$

$$-\log { x } =-2\quad or\quad -\log { x } =-1$$

$$\log { x } =2\quad or\quad \log { x } =1$$

$$x=100\quad or\quad x=10$$

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  • $\begingroup$ your results are ok! $\endgroup$ – Dr. Sonnhard Graubner Oct 4 '15 at 8:57
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    $\begingroup$ Your work is NOT OK because you have no ground in assuming x = 100 in the LS of your 2nd ‘or’ (similarly for letting x= 10 in the RS). Furthermore, if those are true, “going through another 3 steps to say that they are true again” is ‘going round and round’. $\endgroup$ – Mick Oct 4 '15 at 9:12
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    $\begingroup$ @Dr.SonnhardGraubner I guess you are just joking, right? $\endgroup$ – Mick Oct 4 '15 at 9:18
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Hint: Take the logarithm of left and right side of the equation and you'll get a quadratic equation w.r.t. unknown $\log x$.

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  • $\begingroup$ $\log { { x }^{ 3-logx } } =log100\Rightarrow (3-logx)(logx)=2\Rightarrow -(logx)^{ 2 }+3logx-2=0\Rightarrow (logx)^{ 2 }-3logx+2=0\Rightarrow (logx-2)(logx-1)=0\Rightarrow x=100\quad and\quad x=10$ $\endgroup$ – Cherry_Developer Oct 4 '15 at 14:10
  • $\begingroup$ @Cherry_Developer It should be "or" in last formula. Yep, that's exactly what I've meant :) $\endgroup$ – Evgeny Oct 4 '15 at 15:34

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