Question

question 11 (1 point)
evaluate $ln e^5 + ln e^3 + ln e^2$.

a) 10

b) 7

c) 5

d) 4

question 12 (1 point)
the $x$-intercept of the function $y = -4log(x + 3)$ is

a) -4

b) 3

Explanation:

Question 11

Step1: Recall logarithm property

The natural logarithm property states that $\ln e^x = x$ because $\ln$ (natural log) and $e$ are inverse functions. So we apply this to each term:

• For $\ln e^5$, using the property $\ln e^x = x$, we get $5$.
• For $\ln e^3$, using the same property, we get $3$.
• For $\ln e^2$, using the property, we get $2$.

Step2: Sum the results

Now we sum these values: $5 + 3 + 2 = 10$.

Step1: Recall x - intercept definition

The x - intercept of a function $y = f(x)$ is the value of $x$ when $y = 0$. So we set $y = 0$ in the function $y=-4\log(x + 3)$:
$0=-4\log(x + 3)$

Step2: Solve for x

First, divide both sides of the equation $0=-4\log(x + 3)$ by $-4$:
$\frac{0}{-4}=\log(x + 3)$
$0=\log(x + 3)$

We know that if $\log_a b=c$, then $b = a^c$. Assuming the logarithm is base 10 (since it's just written as $\log$), we have:
$x + 3=10^0$
Since $10^0 = 1$, we get:
$x+3 = 1$
Subtract 3 from both sides:
$x=1 - 3=-2$

Wait, there seems to be a mistake in the given options? Wait, maybe the logarithm is natural logarithm? Let's check again. If it's natural logarithm, $\ln(x + 3)=0$, then $x + 3=e^0 = 1$, $x=-2$ still. But the given options are a) - 4, b) 3. Wait, maybe I made a mistake. Wait, let's re - examine the problem.

Wait, the function is $y=-4\log(x + 3)$. Let's set $y = 0$:

$0=-4\log(x + 3)$

Divide both sides by - 4: $\log(x + 3)=0$

If we assume that the logarithm is base 10, then $\log_{10}(x + 3)=0$ implies $x + 3 = 10^0=1$, so $x=1 - 3=-2$. But - 2 is not in the options. Wait, maybe the original function was $y=-4\log(x - 3)$? Or maybe a typo. Wait, if we assume that the function is $y=-4\log(x - 3)$ and set $y = 0$:

$0=-4\log(x - 3)$

$\log(x - 3)=0$

$x - 3=10^0 = 1$

$x=4$. But 4 is not in the options. Wait, maybe the function is $y=-4\log(x+3)$ and we made a mistake. Wait, let's check the options again. The options are a) - 4, b) 3.

Wait, maybe the function is $y = - 4\log(x + 3)$ and we solve it wrong. Wait, if we set $y = 0$:

$0=-4\log(x + 3)$

$\log(x + 3)=0$

$x + 3=1$

$x=-2$. But since - 2 is not in the options, there might be an error in the problem or the options. But according to the given options, maybe there is a mistake in my approach. Wait, maybe the question is about the y - intercept? No, the question says x - intercept.

Wait, maybe the function is $y=-4\log(x + 3)$ and the options are wrong, but if we consider the options given, let's check option b) 3. If $x = 3$, then $y=-4\log(3 + 3)=-4\log(6)
eq0$. Option a) $x=-4$, then $y=-4\log(-4 + 3)=-4\log(-1)$, which is undefined (since the domain of $\log(x)$ is $x>0$). So there is a problem with the given options. But maybe the original function was $y=-4\log(x - 3)$, then when $y = 0$, $\log(x - 3)=0$, $x - 3 = 1$, $x = 4$, still not in options. Or maybe the function is $y=-4\log(x)+3$. Let's try that. Set $y = 0$: $0=-4\log(x)+3$, $4\log(x)=3$, $\log(x)=\frac{3}{4}$, $x = 10^{\frac{3}{4}}\approx5.62$, not in options.

Wait, maybe the question has a typo. But based on the given options, perhaps there is a mistake in the problem statement. However, if we assume that the function is $y=-4\log(x + 3)$ and we made a mistake in the logarithm base. If it's base $e$ (natural log), $\ln(x + 3)=0$, $x+3 = 1$, $x=-2$ still.

Since the given options do not have - 2, there might be an error in the problem or the options. But if we have to choose from the given options, there is a mistake. However, if we re - check the problem, maybe the function is $y=-4\log(x + 3)$ and the options are misprinted. But according to the calculation, the correct x - intercept is $x=-2$, which is not in the options.

Answer:

a) 10

Question 12