Determine $p$ such that $2\cdot \;^3\!\log (5) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1) = \;^3\!\log (p)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$p=20$
Antwoord 2 correct
Fout
Antwoord 3 optie
$p=4$
Antwoord 3 correct
Fout
Antwoord 4 optie
$p=0$
Antwoord 4 correct
Fout
Antwoord 1 optie
$p=10$
Antwoord 1 feedback
Correct: $$\begin{align*}
2\cdot \;^3\!\log (5) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1) & = \;^3\!\log (5^2) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1)\\
& = \;^3\!\log (25) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1)\\
& = \;^3\!\log (\frac{25 \cdot 4 \cdot 1}{10})\\
& = \;^3\!\log (10).\\
\end{align*}$$
Hence, $p=10$.
Go on.
2\cdot \;^3\!\log (5) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1) & = \;^3\!\log (5^2) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1)\\
& = \;^3\!\log (25) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1)\\
& = \;^3\!\log (\frac{25 \cdot 4 \cdot 1}{10})\\
& = \;^3\!\log (10).\\
\end{align*}$$
Hence, $p=10$.
Go on.
Antwoord 2 feedback
Wrong: $\;^3\!\log (25) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1) \neq \;^3\!\log (25-10+4+1)$.
See Properties logarithmic functions.
See Properties logarithmic functions.
Antwoord 3 feedback
Antwoord 4 feedback
Wrong: $\;^3\!\log (25) - \;^3\!\log (10) + \;^3\!\log (4) + \;^3\!\log (1) \neq \;^3\!\log (\frac{10\cdot 4 \cdot 0}{10})$.
See Properties logarithmic functions.
See Properties logarithmic functions.