Determine $p$ and $q$ such that $p=^2\!\log 2^5$ and $q= 3^{^3\!\log 6}$.
$p=5$, $q=6$
$p=32$, $q=729$
$p=25$, $q=216$
$p=10$, $q=18$
Determine $p$ and $q$ such that $p=^2\!\log 2^5$ and $q= 3^{^3\!\log 6}$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$p=32$, $q=729$
Antwoord 2 correct
Fout
Antwoord 3 optie
$p=25$, $q=216$
Antwoord 3 correct
Fout
Antwoord 4 optie
$p=10$, $q=18$
Antwoord 4 correct
Fout
Antwoord 1 optie
$p=5$, $q=6$
Antwoord 1 feedback
Correct:
  1. $^a\!\log a^y$
  2. $x = a^{^a\!\log x}$
Go on.
Antwoord 2 feedback
Wrong: $^2\!\log 2^5\neq 2^5$.

See Relation logarithmic and exponential functions.
Antwoord 3 feedback
Wrong: $^2\!\log 2^5\neq 5^2$.

See Relation logarithmic and exponential functions.
Antwoord 4 feedback
Wrong: $^2\!\log 2^5\neq 2 \cdot 5$.

See Relation logarithmic and exponential functions.