Question 1130719
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            Much more shorter and straightforward solution is possible.



<pre>
First, since "x" is under the logarithm function, it must be positive: x > 0.

So, the domain for this equation is the set {x | x > 0 }.



Second,  {{{10^log(x)}}} = x;  therefore, the original equation in the domain  x > 0  is equivalent to this simple equation


{{{x^4 - 7x^2 + 10}}} = 0.


Factor left side:


(x^2-5)*(x^2-2) = 0.


Formally, it has  4 solutions  {{{sqrt(5)}}},  {{{-sqrt(5)}}},  {{{sqrt(2)}}}  and  {{{-sqrt(2)}}}.


Of these, only positive numbers belong to the domain, so the <U>answer</U> is these two numbers  {{{sqrt(5)}}}  and  {{{sqrt(2)}}}.
</pre>

Solved.


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Regarding the post by &nbsp;@MathLover1, &nbsp;notice that the negative  &nbsp;{{{-sqrt(5)}}}  &nbsp;and  &nbsp;{{{-sqrt(2)}}} &nbsp;<U>are not the solutions</U>.


To count them among the solutions, &nbsp;as &nbsp;@MathLover1 &nbsp;does, &nbsp;is a rude mistake.



The way on how @MathLover1 solves the problem is &nbsp;ANTIPEDAGOGIC.


When the student presents the solution in this way, &nbsp;he &nbsp;(or she) &nbsp;simply demonstrates that &nbsp;he &nbsp;(or she) 
does not know the basic properties of logarithms.



<U>The lesson to learn from my solution is THIS</U> :


<pre>
    the problem in the post is designed and intended to be solved <U>in the way as presented</U>.
</pre>


/\/\/\/\/\/\/\/\/


I don't know for what reason the tutor @greenestamps kicks me and my solution, which is absolutely correct.


Simply ignore/disregard his post in the part which relates to my solution.


In the future, &nbsp;to avoid misunderstanding and ambiguity, &nbsp;write your formulas using parentheses like this:


10^(4*logx) - 7*(10^(2*logx)) + 10 = 0.


It is the plain text format, &nbsp;which is uniquely appropriate format for posting to this forum.


It corresponds to this form of the equation


{{{10^(4*logx)}}} - {{{7*(10^(2*logx))}}} + {{{10}}} = 0.