Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$

Question

Question:

Solve the equations

a) $$\log_{2} x + \log_{3} x = \log_{4} x$$

b) $$\log_{2} x \log_{3} x = \log_{4} x$$

Attempted solution:

The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.

a) $$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$ $$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$

$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$

$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$

Moving the denominator over and solving for $\log_{2} x$

$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$

$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$

b)

$$\log_{2} x \log_{3} x = \log_{4} x$$

$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$

$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$

$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$

Substituting $t = \log_{2} x$ gives:

$$4t^2 - \log_{2}3t = 0 \Rightarrow$$

$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$

$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$

$$t_{1} = 0 \Rightarrow x = 1$$ $$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$

However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?

Answer 1

Your denominator of $4$ in $(b)$ is wrong - $\log_2(4)=2$, not $4$. Then you'd get:

$$2^{\frac{\log_2(3)}{2}} = \left(2^{\log_2(3)}\right)^{1/2}=3^{1/2}$$

It might be easier to just use that $\log_4{x}=\frac{\log_2(x)}{2}$ and solve:

$$\log_2(x)\log_3(x)=\frac{\log_2(x)}{2}$$

This is only true when $\log_2(x)=0$ or when $\log_3(x)=\frac{1}{2}$.

Answer 2

Your mistake in part (b) is on line three where you replaced $\log_24$ with $4$ when it should be replaced by $2$

Answer 3

rewrite the second equation as $$\frac{\ln(x)}{\ln(2)}\cdot\frac{\ln(x)}{\ln(3)}= \frac{\ln(x)}{2\ln(2)}$$ and the first equation as $$\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}=\frac{\ln(x)}{2\ln(2)}$$

Answer 4

One solution is apparent for (a): $x=1$. There can be no other solutions, since the two sides' derivatives are $\frac{1}{x}\left(\frac{1}{\ln(2)}+\frac{1}{\ln(3)}\right)>\frac{1}{x}$ and $\frac{1}{x}\frac{1}{\ln(4)}<\frac{1}{x}$. This shows us that the left side is always growing at a faster rate than the right side, so at most one solution is possible.

For (b), $x=1$ is also an apparent solution. But it's not the only one. Raising $4$ to both sides leaves $$x^{2\log_3(x)}=x\implies x^{2\log_3(x)-1}=1\quad\text{(since $x\ne0$)}$$

So $\ln(x)\cdot(2\log_3(x)-1)=0$. Either $x=1$ to make the first factor $0$, or $2\log_3(x)-1=0$ to make the second factor $0$. The latter option implies $x=3^{1/2}$. So the only two possible solutions are $x=1$ and $x=\sqrt{3}$, both of which check out to actually be solutions.

Answer 5

I think that you made two mistakes.

The first mistake is in a) when factorizing, instead you typed it should have been $\frac{\log_{2} x(2\log_{2} 3+ 2 -\log_{2} 3)}{2\log_{2} 3}=0$ though it doesn´t affect the answer.

And the other mistake in b) is the one that @Gregory Grant says, once you do that you get to answer $x=\sqrt 3$