Question

for each part below, solve the equation. round your answers to the nearest hundredth. do not round any intermediate computations.

(a) solve for (y).
e^y = 3
(y = square)

(b) solve for (x).
7^{x + 5} = 2
(x = square)

Explanation:

Solve the first exponential equation

We begin with the equation in part (a):

$$e^y = 3$$

To solve for \(y\), we apply the natural logarithm (\(\ln\)) to both sides of the equation.
Using the Exponential Functions concept, we know that the natural logarithm is the inverse function of the natural exponential function:

$$\ln(e^y) = \ln(3)$$

$$y = \ln(3)$$

Using a calculator to find the decimal value:

$$y \approx 1.098612...$$

Rounding to the nearest hundredth gives:

$$y \approx 1.10$$

Solve the second exponential equation

Next, we solve the equation in part (b):

$$7^{x+5} = 2$$

To solve for \(x\), we take the logarithm of both sides. We can use the natural logarithm (\(\ln\)):

$$\ln(7^{x+5}) = \ln(2)$$

Using the power property of logarithms, we bring the exponent to the front:

$$(x + 5) \ln(7) = \ln(2)$$

Now, divide both sides by \(\ln(7)\):

$$x + 5 = \frac{\ln(2)}{\ln(7)}$$

Subtract 5 from both sides to isolate \(x\):

$$x = \frac{\ln(2)}{\ln(7)} - 5$$

Using a calculator to compute the value:

$$\frac{\ln(2)}{\ln(7)} \approx \frac{0.693147}{1.945910} \approx 0.356207$$

$$x \approx 0.356207 - 5 = -4.643793...$$

Rounding to the nearest hundredth gives:

$$x \approx -4.64$$

Answer:

Question 1

(a) Solve for \(y\).
\(e^y = 3\)
\(y =\) <blank>\(1.10\)</blank>

Question 2

(b) Solve for \(x\).
\(7^{x+5} = 2\)
\(x =\) <blank>\(-4.64\)</blank>