Breaking Up Logarithms

$\log_x(2x)=2\\ \log_x2+\log_xx=2\\ \log_x2+1=2\\ \log_x2=1\\ 2\log_x2=2\\ \log_x2^2=2\\ \log_x4=2\\ \log_x4+1=2+1\\ \log_x4+\log_xx=3\\ \log_x4x=3$

We know the relation,

if ${a}^{k}=m$ , then $\log _{ a }{ m } =k$

$\therefore \log _{ x }{ 2x } =2\Leftrightarrow { x }^{ 2 }=2x$

$\rightarrow x=2$ or $x=0$

But we can neglect $x=0$ as it doesn't satisfy the given equation.

$\Longrightarrow \log _{ x }{ 4x } =\log _{ 2 }{ 8 } =\log _{ 2 }{ { 2 }^{ 3 } } =\frac { 3\log _{ e }{ 2 } }{ \log _{ e }{ 2 } } =\boxed { 3 }$

$\log_x (2x) = 2$ , we have $2x = x^2$ , this gives us $x=2$ . $\log_x (4x)=\log_2 (4x)=\log_2 (2^3)=3$ .

First, you write $\log_x 2x$ = 2 into exponential form:

$x^{2}$ = 2x, which by solving out, we get x = 2 or x = 0. We will have to ignore x = 0, though since x = 0 does not satisfy the original equation.

Next, you write $\log_x 4x$ = y into exponential form:

$x^{y}$ = 4x.

By plugging in our found x value, we now get:

$2^{y}$ = 8.

What we do know about this relationship is that 8 is equal to $2^{3}$ , so y must equal 3.

Therefore, our final solution must be $\boxed{3}$

$log_x(2x)=2$ $\implies$ $x^2=2x$

Either $0$ or $2$ satisfies this.

$log_0(0)= undefined$ (Substituting $0$ in place of $x$ in $log_x(4x)$ )

$\therefore$

$log_x(4x)=log_2(8)=3$ (Substituting $2$ in place of $x$ in $log_x(4x)$ )

Hence $3$ is our answer.