Exercise 2 | Wiskunde op Tilburg University

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.

Antwoord 2 feedback

Wrong: The solution to

$4=\;^2\!\log x$ is not

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.

Antwoord 4 feedback

Wrong:

$(\;^2\!\log x)^2 \neq \;^2\!\log (x^2)$ .

See Properties logarithmic functions.