Solve $(\;^2\!\log x)^2+12=7\cdot \;^2\!\log x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x=16$ or $x=9$
Antwoord 2 correct
Fout
Antwoord 3 optie
$x=3$ or $x=4$
Antwoord 3 correct
Fout
Antwoord 4 optie
$x=2^{\frac{12}{5}}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$x=16$ or $x=8$
Antwoord 1 feedback
Correct: We define $p=\;^2\!\log x$. Then we obtain
$$\begin{align*}
p^2+12=7\cdot p & \Leftrightarrow p^2-7p+12=0\\
& \Leftrightarrow (p-3)(p-4)\\
& \Leftrightarrow p=3 \mbox{ of } p=4.
\end{align*}$$
$3=\;^2\!\log x \Leftrightarrow x=2^3=8$ and $4=\;^2\!\log x \Leftrightarrow x=2^4=16$.
Go on.
$$\begin{align*}
p^2+12=7\cdot p & \Leftrightarrow p^2-7p+12=0\\
& \Leftrightarrow (p-3)(p-4)\\
& \Leftrightarrow p=3 \mbox{ of } p=4.
\end{align*}$$
$3=\;^2\!\log x \Leftrightarrow x=2^3=8$ and $4=\;^2\!\log x \Leftrightarrow x=2^4=16$.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Fout: $(\;^2\!\log x)^2+12=7\cdot \;^2\!\log x$ is not equal to $x^2+12=7x$.
See Logarithmic functions.
See Logarithmic functions.
Antwoord 4 feedback