Properties of Logarithm (what is wrong with my math?)

Question

The question is to find the intersection point between $y = log_22x$ and $y = log_4x$

So my first instinct was to make the same base 2. (Since 4 = 2 + 2)

$log_22x = log_4x$

$log_22 + log_2x = log_4x$

$1 + log_2x = log_4x$

$1 +\frac{1}{log_x2} = \frac{1}{log_x4}$

$1 +\frac{1}{log_x2} = \frac{1}{log_x2 + log_x2}$

Let y = $log_x2$

$1 + \frac{1}{y} = \frac{1}{y+y}$

$\frac{y+1}{y} = \frac{1}{y+y}$

Cross multiply

$(y+1)(y+y) = (y)(1)$

$2y^2 + 2y = y$

$2y^2 + y = 0$

$y(2y + 1) = 0$

$y = 0$ or $y =-1/2$ (won't work since argument has to be greater than 0)

What is wrong with my math???

How would you tackle this problem? Your instinct? (I want to know various ways to tackle this problem because I gave up on the first one but got it right using this one:

$log_22x = log_4x$

$4^{log_22x} = x$

$2^{log_22x} * 2^{log_22x} = x$

$2x * 2x = x$

$4x^{2} - x = 0$

$x = 0, x= 1/4$

x = 1/4 is the right answer

THIS IS PRE-CALCULUS QUESTION (Please explain at a pre-cal level Thanks)

Answer 1

In fact, your first method is perfectly correct! There's no reason to discard the $y=-\frac12$ answer because $y$ is not the argument of a logarithm (so needs not be positive). Remembering that $y=\log_x2$, we compute $x$:

$$-\frac12 = \log_x2$$ $$x^{-\frac12} = 2$$ $$x = (x^{-\frac12})^{-2} = 2^{-2} = \frac14$$ which matches the answer you got from the other method.

Answer 2

You did nothing wrong, except you did not finish. $y=-1/2$ means $\log_x 2=-1/2$ or $\log_2 x=-2$ or $x=1/4$

Answer 3

Nothing wrong, but…

In general, $\log_a(x)=\dfrac{\log_b(x)}{\log_b(a)}$. Indeed, if $y=\log_a(x)$, then $x=a^y$ and therefore $\log_b(x)=y\log_b(a)$. Thus $$ \log_a(x)=y=\dfrac{\log_b(x)}{\log_b(a)} $$

With this at hand, the equation $$ \log_2(2x)=\log_4(x) $$ can be rewritten as $$ \log_2(2x)=\frac{\log_2(x)}{\log_2(4)} $$ that, using $\log_2(4)=2$, simplifies to $2\log_2(2x)=\log_2(4)$, that is, $$ \log_2(4x^2)=\log_2(x) $$ This forces $4x^2=x$ and, being $x>0$ by assumption, $x=1/4$.