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I'm supposed to solve for $x$ and find the domain. I know that adding two logs together is multiplication of the numbers, but what if two logs are completely multiplying each other? $$\log_{10}(0.1x^2)\log_{10}x=1$$

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  • $\begingroup$ What do you mean by "find the domain"? $\endgroup$ – Eric Wofsey Nov 4 '18 at 1:07
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    $\begingroup$ Possible hint. Since $\log(x^2) = 2\log(x)$ you can rewrite this as a quadratic equation in $\log(x)$, solve for that and then for $x$. $\endgroup$ – Ethan Bolker Nov 4 '18 at 1:10
  • $\begingroup$ The domain of the left side is $x>0$ [input to any log needs to be positive, so first log restricts to nonzero $x$ as it's squared, but second log restricts $x$ to be positive. $\endgroup$ – coffeemath Nov 4 '18 at 1:10
  • $\begingroup$ Not much (as far as what you've asked is concerned): there is the identity $\log a\log b=\log (b^{\log a})$ for $a,b> 0$, but that's not quite the way you were meant to face the exercise. $\endgroup$ – Saucy O'Path Nov 4 '18 at 1:11
  • $\begingroup$ $log_{10}(0.1x^2)log_{10}(x)=(-1+2log_{10}(x))log_{10}(x)=2log_{10}^2(x)-log_{10}(x)=1$ $2log_{10}^2(x)=log_{10}(10)+log_{10}(x)$ $2log_{10}^2(x)=log_{10}(10x)$ Beyond that, I'm not really sure, but hopefully that is helpful $\endgroup$ – Seth Nov 4 '18 at 1:16
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The domain of the LHS of the original equation is $x\in(0,\infty)$, because of the presence of $\log_{10}x$.

By logarithm rules (justified by the domain of $x$), we can take the multiple of 0.1 and the square out of $0.1x^2$: $$\log_{10}(0.1x^2)\log_{10}x=(\log_{10}0.1+2\log_{10}x)\log_{10}x=(2\log_{10}x-1)\log_{10}x=1$$ We denote $\log_{10}x=y$: $$(2y-1)y=2y^2-y=1$$ $$2y^2-y-1=0$$ Solving for $y$, we get $y=1$ and $y=-\frac12$, corresponding to solutions of $x=10$ and $x=\frac1{\sqrt{10}}=0.316\dots$

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  • $\begingroup$ While it is true that $2\log_{10} x = \log_{10} x^2$ for $x > 0$, the domain of $\log_{10} x^2$ is $(-\infty, 0) \cup (0, \infty)$ since $x^2 > 0$ unless $x = 0$. Hence, $\log_{10} x^2 = 2\log_{10} |x|$, which you will notice also has domain $(-\infty, 0) \cup (0, \infty)$. $\endgroup$ – N. F. Taussig Nov 4 '18 at 11:36
  • $\begingroup$ @N.F.Taussig But there's still a lone $\log_{10}x$ in the original equation. $\endgroup$ – Parcly Taxel Nov 4 '18 at 11:36
  • $\begingroup$ I am aware of that. What I am saying does not change the answer. Perhaps it would be better to point out that the domain of the expression $\log_{10} (0.1x^2)\log_{10} x$ at the start of the answer rather than at the end, which would then justify your assertion that $\log_{10} x^2 = 2\log_{10} x$. $\endgroup$ – N. F. Taussig Nov 4 '18 at 11:40
  • $\begingroup$ @N.F.Taussig done. $\endgroup$ – Parcly Taxel Nov 4 '18 at 11:44

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