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$$\sqrt{\log_x\left(\sqrt{3x}\right)} \cdot \log_3 x = -1$$

I am not entirely sure how to go about solving for $x$. I cannot square each side because the product isn't $≥ 0$, I can't think of any more approaches right now.

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Where does the first square root end? –  Frunobulax Aug 14 at 21:06
Use { and } braces to properly define subexpressions for radicals. –  CiaPan Aug 14 at 21:16
You might divide both sides by $\log_3x$, then use the identity $$\log_ba = \frac{\log_ca}{\log_cb}$$ which implies $$\log_ba = \frac 1{\log_ab}$$ –  CiaPan Aug 14 at 21:44

3 Answers 3

up vote 1 down vote accepted

Will things look better without logarithms in the equation? First, let's simplify $\log_x\sqrt{3x}$.


Now, we'll use the info from Ciapan's comment: $\log_x3=\dfrac1{\log_3x}$. Let $y=\log_x3$. The equation now becomes


Can you solve that? You may want to check your answers after to make sure they make sense.

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Since square roots are nonnegative, we know that $\log_3 x < 0$ so that $0 < x < 1$. Now let $k = \log 3$ and let $y = \log x$. Then by using log rules, our equation becomes: \begin{align*} \sqrt{\frac{\log \sqrt{3x}}{\log x}} \cdot \frac{\log x}{\log 3} &= -1 \\ \sqrt{\frac{\frac{1}{2}\log 3x}{\log x}} \cdot \frac{\log x}{\log 3} &= -1 \\ \sqrt{\frac{\frac{1}{2}(\log 3 + \log x)}{\log x}} &= -\frac{\log 3}{\log x} \\ \sqrt{\frac{\frac{1}{2}(k + y)}{y}} &= -\frac{k}{y} \\ \frac{\frac{1}{2}(k + y)}{y} &= \frac{k^2}{y^2} \\ ky + y^2 &= 2k^2 \\ y^2 + ky + \frac{1}{4}k^2 &= \frac{9}{4}k^2 \\ \left(y + \frac{1}{2}k\right)^2 &= \frac{9}{4}k^2 \\ y + \frac{1}{2}k &= \pm \frac{3}{2}k \\ y &= k,-2k \\ \end{align*} Converting back, we find that either $\log x = \log 3$ or $\log x = -2\log 3 = \log (1/9)$ so that either $x = 3$ or $x = 1/9$. But then since $0 < x < 1$, we reject the first extraneous solution and conclude that $x = 1/9$.

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Isn't $\log(3)>0$? –  cjferes Aug 14 at 22:01
On line six, where you square each side - I don't think you have the mathematical right to do that. For example f = g , f^2 = g^2 iff f*g>0. Using your logic, squaring each side of 3 = -3 would result in 9 = 9. You can derive anything by improper logic, thus this is very dangerous. –  user142795 Aug 14 at 22:04
@Adriano that's not right. $\log(x)<0\leftrightarrow x<1$ and it doesn't matter what the base is. You are thinking of $\log_x(3)<1$, and that clearly dependes on the base $x$ of the logarithm. –  cjferes Aug 14 at 22:06
Squaring both sides of an equation is always legal ($3 = -3 \implies 9 = 9$ is vacuously true because the antecedent is false). Square rooting both sides of an equation is not always legal (for example, $9 = 9 \implies 3 = -3$ is false), and can be avoided by using a plus-or-minus symbol and then checking at the end for extraneous solutions. –  Adriano Aug 14 at 22:06
They only need to have the same sign when you square root both sides (if $fg > 0$ and if $f^2 = g^2$, then $f = g$). If we are squaring both sides, then this extra restriction is unnecessary (if $f = g$, then $f^2 = g^2$). Note that squaring and square rooting are not truly inverses of each other. If you want these operations to be reversible, then you'll need to add the extra condition of having the same sign. –  Adriano Aug 14 at 22:19

$$\log_x(\sqrt{3x}) = \log_x(3x)^{\frac{1}{2}} = \frac{1}{2} \log_x(3x)=(*)$$

To change the base of the logarithm, so that you have at both logarithms the base $3$, use the formula: $$\log_b a=\frac{\log_c a}{\log_c b}$$

$$\log_x {3x}=\frac{\log_3 {3x}}{\log_3 {x}}=\frac{\log_3 {3}+\log_3 {x}}{\log_3 {x}}=\frac{1+\log_3 {x}}{\log_3 {x}}$$

$$(*)=\frac{1}{2} \frac{1+\log_3 {x}}{\log_3 {x}}=\frac{1}{2} \left ( \frac{1}{\log_3{x}}+1 \right )$$

$$\sqrt{\log_x(\sqrt{3x})} \cdot \log_3{x}=-1 \Rightarrow \sqrt{\frac{1}{2} \left ( \frac{1}{\log_3{x}}+1 \right )} \cdot \log_3{x}=-1 \\ \Rightarrow \sqrt{\frac{\log_3^2{x}}{2} \left ( \frac{1}{\log_3{x}}+1 \right )} =-1 \Rightarrow \sqrt{\frac{1}{2} \left ( \log_3{x}+\log_3^2{x} \right )} =\sqrt{i} \Rightarrow \frac{1}{2} \left ( \log_3{x}+\log_3^2{x} \right ) =i $$

Continue by solving for $x$.

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