AHL 1.9—Log laws

The equation $\log 5+\log 20=\log c$ matches the product rule: ${\log }_a(x)+{\log }_a(y)={\log }_a(xy)$, since both logarithms share base 10.

$\log 5+\log 20=\log (5\times 20)=\log (100)$

Since $\log (100)=\log c$, we have $c=100$. This is consistent with $\log (100)=2$ since $10^2=100$.

We need to find $c$ such that $\log c=\log 5+\log 20$. The product rule tells us $c=5\times 20$.

$c=25$ would correspond to $5+20=25$, which confuses addition of the numbers inside the logs with the product rule. Eliminated.

$c=15$ has no logical connection to combining $\log 5$ and $\log 20$ under any valid log law. Eliminated.

$c=1000$ would mean $\log c=3$, but $\log 5+\log 20=\log (100)=2\ne 3$. Eliminated.

$c=100$ because $5\times 20=100$, and $\log (100)=2$, which matches $\log 5+\log 20\approx 0.699+1.301=2$. ✓

By definition, ${\log }_{3}81=k$ means $3^k=81$. We need to find the power of 3 that gives 81.

$81=3^4\text{since }{3}^1=3,3^2=9,3^3=27,3^4=81$

Therefore $k=4$. This can also be confirmed using the power rule: ${\log }_3(3^4)=4{\log }_3(3)=4\times 1=4$.

We must find integer $k$ such that $3^k=81$. Test each option.

$3^2=9\ne 81$. Eliminated.

$3^3=27\ne 81$. Eliminated.

$3^9=19683\ne 81$. Eliminated.

$3^4=81$, so $k=4$. ✓