Determine all $x$ such that $ 3\cdot \;^5\!\log (x)<\;^5\!\log (5x)+\;^5\!\log (60x) -(\;^5\!\log(3x)+2)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x<-2$ and $0<x<2$
Antwoord 2 correct
Fout
Antwoord 3 optie
$x>0$
Antwoord 3 correct
Fout
Antwoord 4 optie
$x<-2$ and $x>2$
Antwoord 4 correct
Fout
Antwoord 1 optie
$0<x<2$
Antwoord 1 feedback
Correct: $$\begin{align*}
3\cdot \;^5\!\log (x)=\;^5\!\log (5x)+\;^5\!\log (60x) -(\;^5\!\log(3x)+2)\\
\Leftrightarrow &\;^5\!\log (x^3)=\;^5\!\log (5x)+\;^5\!\log (60x) -\;^5\!\log(3x)-\;^5\!\log (25)\\
\Leftrightarrow &\;^5\!\log (x^3)=\;^5\!\log (\frac{5x\cdot 60x}{3x \cdot 25})\\
\Leftrightarrow & \;^5\!\log (x^3)=\;^5\!\log (4x)\\
\Leftrightarrow &x^3=4x\\
\Leftrightarrow &x^3-4x=0\\
\Leftrightarrow &x(x^2-4)=0\\
\Leftrightarrow & x=0 \mbox{ or } x=-2 \mbox{ or } x=2.
\end{align*}$$
$x=0$ and $x=-2$ are outside the domain of the function. Hence, $x=2$. Via a sign chart we find $0<x<2$.
Go on.
3\cdot \;^5\!\log (x)=\;^5\!\log (5x)+\;^5\!\log (60x) -(\;^5\!\log(3x)+2)\\
\Leftrightarrow &\;^5\!\log (x^3)=\;^5\!\log (5x)+\;^5\!\log (60x) -\;^5\!\log(3x)-\;^5\!\log (25)\\
\Leftrightarrow &\;^5\!\log (x^3)=\;^5\!\log (\frac{5x\cdot 60x}{3x \cdot 25})\\
\Leftrightarrow & \;^5\!\log (x^3)=\;^5\!\log (4x)\\
\Leftrightarrow &x^3=4x\\
\Leftrightarrow &x^3-4x=0\\
\Leftrightarrow &x(x^2-4)=0\\
\Leftrightarrow & x=0 \mbox{ or } x=-2 \mbox{ or } x=2.
\end{align*}$$
$x=0$ and $x=-2$ are outside the domain of the function. Hence, $x=2$. Via a sign chart we find $0<x<2$.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: Calculate all the solutions of the corresponding equation.
See Properties logarithmic functions.
See Properties logarithmic functions.
Antwoord 4 feedback