Logarithm Equality

Question

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

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.

Answer 1

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

$$\log_x\sqrt{3x}=\frac12\log_x3x=\frac12(\log_x3+\log_xx)=\frac12(\log_x3+1)$$

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

$$\dfrac{\sqrt{\frac12(y+1)}}y=-1$$

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

Answer 2

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

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 $x$.

Answer 3

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 $x = 1/9$.