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I have this exercise:

Solve for $x$, $(\log x)^{10} = 3$

My development was:

$\log x = \sqrt[10]3$

$10^{\sqrt[10]3} = x$ , this is my solution.

But I still need a solution, according to the symbolab. Which is: $x = \frac{1}{10^{3^{\frac{1}{10}}}}$

But I have not managed to get it on my own, how can I get to this result? Thanks in advance.

Bernard
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ESCM
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    In real numbers $,z^{10}=3,$ has the roots $,z = \color{red}{\pm} \sqrt[10]{3},$. – dxiv Aug 20 '18 at 22:20
  • How you know that? – ESCM Aug 20 '18 at 22:24
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    $z^{10} = (-z)^{10},$, so if $,z,$ is a root then $,-z,$ is also a root. As for why it does not have more than two real roots, think at the convexity of $,z^{10},$ for example. – dxiv Aug 20 '18 at 22:28

1 Answers1

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Let's try this with a much simpler problem: suppose you want to find $y$ such that $y^{10} = 1.$

Obviously $y = \sqrt[10]1 = 1$ is a solution.

Is there any other solution? Why or why not?

Your solution method has much in common with the method for finding that $y = 1$ in the example above: you have something to the $10$th power, so you assume all you have to do is take the $10$th root. But this does not give complete solutions, just as taking the square root does not completely solve equations like $x^2 = 1.$

David K
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  • True, so how i can get all solutions in this type of exercise? – ESCM Aug 20 '18 at 22:25
  • One way is to remember that raising any number to an even power will give it a positive sign, even if the original number was negative. So working backward, when you take the $n$th root, you have to consider both possibilities, that you had something positive or something negative before taking the $n$th power. – David K Aug 20 '18 at 22:56