Why does it for $x^2=9$ we get two solutions, while if we use the "log both sides" property the negative solution is rejected? which method is true and why?
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1When you write $\log(x^2)=2\log x$ you are assuming $x>0$. – Intelligenti pauca Jan 07 '16 at 21:36
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The property $\log(a^n)=n \log(a)$ is true if $a>0$ and $n \in \mathbb N$, but not if $a<0$, since $\log$ is only defined on positive real numbers.
Therefore, using $\log$ is not the right method. You should better use $x^2-9=(x-3)(x+3)$ which holds for all $x \in \mathbb R$.
Watson
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1The logarithm method works fine with this equation. You just have to be careful. The correct application would say that "$x^2=9$" is equivalent to "$2\log x = \log 9$ if $x>0$, or $2\log(-x)=\log 9$ if $x<0$". It's because $x^2$ and $(-x)^2$ are the same number. – MPW Jan 07 '16 at 21:49
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Because $\log_3 x^2=2\log_3|x|$, so from $x^2=9$ you get $$ 2\log_3|x|=\log_39=2 $$ and so $$ \log_3|x|=1 $$ from which $|x|=3$.
Be careful: $\sqrt{x^2}=|x|$.
egreg
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