Question Solve $\ln c = 15$ and write the exact solution set. Also provide approximate solutions to 4 decimal places.

Studdy Solution

STEP 1

Assumptions1. The given equation is $\ln c =15$ . We are solving for $c$
3. We are asked to provide both the exact solution and an approximate solution to4 decimal places

STEP 2

To solve for $c$ , we need to get rid of the natural logarithm on the left side of the equation. We can do this by applying the property of logarithms that states $e^{\ln a} = a$ .So, we take the exponential of both sides of the equation.
$e^{\ln c} = e^{15}$

STEP 3

implify the left side of the equation using the property of logarithms mentioned above.
$c = e^{15}$ This is the exact solution.

STEP 4

Now, to find the approximate solution to4 decimal places, we need to calculate the value of $e^{15}$ .

STEP 5

Calculate the value of $e^{15}$ to4 decimal places.
$c \approx e^{15} \approx3269017.3725$ So, the solution set is $c = e^{15}$ or approximated to4 decimal places is $c \approx3269017.3725$ .

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Question Solve $\ln c = 15$ and write the exact solution set. Also provide approximate solutions to 4 decimal places.

Studdy Solution

STEP 1

Assumptions1. The given equation is $\ln c =15$ . We are solving for $c$
3. We are asked to provide both the exact solution and an approximate solution to4 decimal places

STEP 2

To solve for $c$ , we need to get rid of the natural logarithm on the left side of the equation. We can do this by applying the property of logarithms that states $e^{\ln a} = a$ .So, we take the exponential of both sides of the equation.
$e^{\ln c} = e^{15}$

STEP 3

implify the left side of the equation using the property of logarithms mentioned above.
$c = e^{15}$ This is the exact solution.

STEP 4

Now, to find the approximate solution to4 decimal places, we need to calculate the value of $e^{15}$ .

STEP 5

Calculate the value of $e^{15}$ to4 decimal places.
$c \approx e^{15} \approx3269017.3725$ So, the solution set is $c = e^{15}$ or approximated to4 decimal places is $c \approx3269017.3725$ .

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Question Solve ln⁡c=15\ln c = 15lnc=15 and write the exact solution set. Also provide approximate solutions to 4 decimal places.

Studdy Solution

STEP 1

Assumptions1. The given equation is ln⁡c=15\ln c =15lnc=15 . We are solving for ccc3. We are asked to provide both the exact solution and an approximate solution to4 decimal places

STEP 2

To solve for ccc, we need to get rid of the natural logarithm on the left side of the equation. We can do this by applying the property of logarithms that states eln⁡a=ae^{\ln a} = aelna=a.So, we take the exponential of both sides of the equation.eln⁡c=e15e^{\ln c} = e^{15}elnc=e15

STEP 3

implify the left side of the equation using the property of logarithms mentioned above.c=e15c = e^{15}c=e15This is the exact solution.

STEP 4

Now, to find the approximate solution to4 decimal places, we need to calculate the value of e15e^{15}e15.

STEP 5

Calculate the value of e15e^{15}e15 to4 decimal places.c≈e15≈3269017.3725c \approx e^{15} \approx3269017.3725c≈e15≈3269017.3725So, the solution set is c=e15c = e^{15}c=e15 or approximated to4 decimal places is c≈3269017.3725c \approx3269017.3725c≈3269017.3725.

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Studdy solves anything!

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question Solve $\ln c = 15$ and write the exact solution set. Also provide approximate solutions to 4 decimal places.

Assumptions1. The given equation is $\ln c =15$ . We are solving for $c$
3. We are asked to provide both the exact solution and an approximate solution to4 decimal places

To solve for $c$ , we need to get rid of the natural logarithm on the left side of the equation. We can do this by applying the property of logarithms that states $e^{\ln a} = a$ .So, we take the exponential of both sides of the equation.
$e^{\ln c} = e^{15}$

implify the left side of the equation using the property of logarithms mentioned above.
$c = e^{15}$ This is the exact solution.

Now, to find the approximate solution to4 decimal places, we need to calculate the value of $e^{15}$ .

Calculate the value of $e^{15}$ to4 decimal places.
$c \approx e^{15} \approx3269017.3725$ So, the solution set is $c = e^{15}$ or approximated to4 decimal places is $c \approx3269017.3725$ .