SL 2.9—Exponential and logarithmic functions

We need to evaluate $N(t)$ at $t=0$ to find the initial number of bacteria.

$N(0)=800\ln (0+2)=800\ln 2$

$\ln 2\approx 0.693$, so $N(0)\approx 800\times 0.693\approx 554$ bacteria. The exact answer is $800\ln 2$.

The answer is $800\ln 2$.

We substitute $t=0$ into $N(t)=800\ln (t+2)$ and check each option.

$800$ would be the answer only if $\ln (0+2)=1$, i.e. $\ln 2=1$, which is false since $\ln 2\approx 0.693$. Eliminated.

$0$ would require $\ln (t+2)=0$, meaning $t+2=1$, so $t=-1$, not $t=0$. Eliminated.

$800\ln 1=800\times 0=0$. This corresponds to substituting $t=-1$, not $t=0$. Eliminated.

$800\ln 2$ is correct because $N(0)=800\ln (0+2)=800\ln 2$.

The argument of a logarithm must be strictly positive. So we need $x^2-9>0$.

$x^2-9>0⟹(x-3)(x+3)>0$

The product $(x-3)(x+3)>0$ when both factors are positive ($x>3$) or both are negative ($x<-3$).

Domain: $x<-3$ or $x>3$.

We need $x^2-9>0$ for the logarithm to be defined.

This ignores the left branch. For example, $x=-4$ gives $(-4{)}^2-9=7>0$, so $x=-4$ should be in the domain. This option is incomplete. Eliminated.

Testing $x=0$: $0^2-9=-9<0$, which is not valid for a logarithm. This interval makes the argument negative. Eliminated.

This would include values like $x=0$ where $x^2-9=-9<0$, which is invalid. Eliminated.

$x<-3$ or $x>3$ correctly captures all $x$ where $x^2-9>0$.