# How do you multiply these logarithms and find the domain?

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? $$\log_{10}(0.1x^2)\log_{10}x=1$$

• What do you mean by "find the domain"? – Eric Wofsey Nov 4 '18 at 1:07
• 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 $x$. – Ethan Bolker Nov 4 '18 at 1:10
• 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. – coffeemath Nov 4 '18 at 1:10
• 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. – Saucy O'Path Nov 4 '18 at 1:11
• $log_{10}(0.1x^2)log_{10}(x)=(-1+2log_{10}(x))log_{10}(x)=2log_{10}^2(x)-log_{10}(x)=1$ $2log_{10}^2(x)=log_{10}(10)+log_{10}(x)$ $2log_{10}^2(x)=log_{10}(10x)$ Beyond that, I'm not really sure, but hopefully that is helpful – Seth Nov 4 '18 at 1:16

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 $$x=\frac1{\sqrt{10}}=0.316\dots$$

• 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 $(-\infty, 0) \cup (0, \infty)$. – N. F. Taussig Nov 4 '18 at 11:36
• @N.F.Taussig But there's still a lone $\log_{10}x$ in the original equation. – Parcly Taxel Nov 4 '18 at 11:36
• 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 $\log_{10} x^2 = 2\log_{10} x$. – N. F. Taussig Nov 4 '18 at 11:40
• @N.F.Taussig done. – Parcly Taxel Nov 4 '18 at 11:44