Solved on Dec 09, 2023

Solve the equation $7e^{2m-1} - 3 = 11$ . Find the exact solution set using logarithms and approximate solutions to 4 decimal places.

STEP 1

Assumptions
1. The given equation is $7 e^{2 m-1}-3=11$ .
2. We need to solve for $m$ .
3. The solution should be expressed in terms of common or natural logarithms.
4. An approximate solution should be provided to at least 4 decimal places.

STEP 2

First, we need to isolate the exponential term on one side of the equation. To do this, we will add $3$ to both sides of the equation.
$7 e^{2 m-1} - 3 + 3 = 11 + 3$

STEP 3

Simplify both sides of the equation.
$7 e^{2 m-1} = 14$

STEP 4

Next, we will divide both sides of the equation by $7$ to solve for the exponential term $e^{2 m-1}$ .
$\frac{7 e^{2 m-1}}{7} = \frac{14}{7}$

STEP 5

Simplify both sides of the equation.
$e^{2 m-1} = 2$

STEP 6

Now, we will take the natural logarithm of both sides of the equation to solve for $2m - 1$ .
$\ln(e^{2 m-1}) = \ln(2)$

STEP 7

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the left side of the equation.
$2 m - 1 = \ln(2)$

STEP 8

Next, we will add $1$ to both sides of the equation to solve for $2m$ .
$2 m - 1 + 1 = \ln(2) + 1$

STEP 9

Simplify both sides of the equation.
$2 m = \ln(2) + 1$

STEP 10

Now, we will divide both sides of the equation by $2$ to solve for $m$ .
$m = \frac{\ln(2) + 1}{2}$

STEP 11

This is the exact solution in terms of natural logarithms. We can now find an approximate solution by evaluating the right side of the equation using a calculator.
$m \approx \frac{\ln(2) + 1}{2}$

STEP 12

Using a calculator, we find the approximate value of $\ln(2)$ .
$\ln(2) \approx 0.6931$

STEP 13

Substitute the approximate value of $\ln(2)$ into the equation.
$m \approx \frac{0.6931 + 1}{2}$

STEP 14

Calculate the approximate value of $m$ .
$m \approx \frac{1.6931}{2}$

STEP 15

Simplify the approximation.
$m \approx 0.84655$

STEP 16

Round the approximation to at least 4 decimal places.
$m \approx 0.8466$
The exact solution set is $\left\{\frac{\ln(2) + 1}{2}\right\}$ . $m \approx 0.8466$

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Solve the equation $7e^{2m-1} - 3 = 11$ . Find the exact solution set using logarithms and approximate solutions to 4 decimal places.

STEP 1

Assumptions
1. The given equation is $7 e^{2 m-1}-3=11$ .
2. We need to solve for $m$ .
3. The solution should be expressed in terms of common or natural logarithms.
4. An approximate solution should be provided to at least 4 decimal places.

STEP 2

First, we need to isolate the exponential term on one side of the equation. To do this, we will add $3$ to both sides of the equation.
$7 e^{2 m-1} - 3 + 3 = 11 + 3$

STEP 3

Simplify both sides of the equation.
$7 e^{2 m-1} = 14$

STEP 4

Next, we will divide both sides of the equation by $7$ to solve for the exponential term $e^{2 m-1}$ .
$\frac{7 e^{2 m-1}}{7} = \frac{14}{7}$

STEP 5

Simplify both sides of the equation.
$e^{2 m-1} = 2$

STEP 6

Now, we will take the natural logarithm of both sides of the equation to solve for $2m - 1$ .
$\ln(e^{2 m-1}) = \ln(2)$

STEP 7

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the left side of the equation.
$2 m - 1 = \ln(2)$

STEP 8

Next, we will add $1$ to both sides of the equation to solve for $2m$ .
$2 m - 1 + 1 = \ln(2) + 1$

STEP 9

Simplify both sides of the equation.
$2 m = \ln(2) + 1$

STEP 10

Now, we will divide both sides of the equation by $2$ to solve for $m$ .
$m = \frac{\ln(2) + 1}{2}$

STEP 11

This is the exact solution in terms of natural logarithms. We can now find an approximate solution by evaluating the right side of the equation using a calculator.
$m \approx \frac{\ln(2) + 1}{2}$

STEP 12

Using a calculator, we find the approximate value of $\ln(2)$ .
$\ln(2) \approx 0.6931$

STEP 13

Substitute the approximate value of $\ln(2)$ into the equation.
$m \approx \frac{0.6931 + 1}{2}$

STEP 14

Calculate the approximate value of $m$ .
$m \approx \frac{1.6931}{2}$

STEP 15

Simplify the approximation.
$m \approx 0.84655$

STEP 16

Round the approximation to at least 4 decimal places.
$m \approx 0.8466$
The exact solution set is $\left\{\frac{\ln(2) + 1}{2}\right\}$ . $m \approx 0.8466$

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Solve the equation 7e2m−1−3=117e^{2m-1} - 3 = 117e2m−1−3=11. Find the exact solution set using logarithms and approximate solutions to 4 decimal places.

STEP 1

Assumptions1. The given equation is 7e2m−1−3=117 e^{2 m-1}-3=117e2m−1−3=11.2. We need to solve for mmm.3. The solution should be expressed in terms of common or natural logarithms.4. An approximate solution should be provided to at least 4 decimal places.

STEP 2

First, we need to isolate the exponential term on one side of the equation. To do this, we will add 333 to both sides of the equation.7e2m−1−3+3=11+37 e^{2 m-1} - 3 + 3 = 11 + 37e2m−1−3+3=11+3

STEP 3

Simplify both sides of the equation.7e2m−1=147 e^{2 m-1} = 147e2m−1=14

STEP 4

Next, we will divide both sides of the equation by 777 to solve for the exponential term e2m−1e^{2 m-1}e2m−1.7e2m−17=147\frac{7 e^{2 m-1}}{7} = \frac{14}{7}77e2m−1​=714​

STEP 5

Simplify both sides of the equation.e2m−1=2e^{2 m-1} = 2e2m−1=2

STEP 6

Now, we will take the natural logarithm of both sides of the equation to solve for 2m−12m - 12m−1.ln⁡(e2m−1)=ln⁡(2)\ln(e^{2 m-1}) = \ln(2)ln(e2m−1)=ln(2)

STEP 7

Using the property of logarithms that ln⁡(ex)=x\ln(e^x) = xln(ex)=x, we can simplify the left side of the equation.2m−1=ln⁡(2)2 m - 1 = \ln(2)2m−1=ln(2)

STEP 8

Next, we will add 111 to both sides of the equation to solve for 2m2m2m.2m−1+1=ln⁡(2)+12 m - 1 + 1 = \ln(2) + 12m−1+1=ln(2)+1

STEP 9

Simplify both sides of the equation.2m=ln⁡(2)+12 m = \ln(2) + 12m=ln(2)+1

STEP 10

Now, we will divide both sides of the equation by 222 to solve for mmm.m=ln⁡(2)+12m = \frac{\ln(2) + 1}{2}m=2ln(2)+1​

STEP 11

This is the exact solution in terms of natural logarithms. We can now find an approximate solution by evaluating the right side of the equation using a calculator.m≈ln⁡(2)+12m \approx \frac{\ln(2) + 1}{2}m≈2ln(2)+1​

STEP 12

Using a calculator, we find the approximate value of ln⁡(2)\ln(2)ln(2).ln⁡(2)≈0.6931\ln(2) \approx 0.6931ln(2)≈0.6931

STEP 13

Substitute the approximate value of ln⁡(2)\ln(2)ln(2) into the equation.m≈0.6931+12m \approx \frac{0.6931 + 1}{2}m≈20.6931+1​

STEP 14

Calculate the approximate value of mmm.m≈1.69312m \approx \frac{1.6931}{2}m≈21.6931​

STEP 15

Simplify the approximation.m≈0.84655m \approx 0.84655m≈0.84655

STEP 16

Round the approximation to at least 4 decimal places.m≈0.8466m \approx 0.8466m≈0.8466The exact solution set is {ln⁡(2)+12}\left\{\frac{\ln(2) + 1}{2}\right\}{2ln(2)+1​}. m≈0.8466 m \approx 0.8466 m≈0.8466

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Solve the equation $7e^{2m-1} - 3 = 11$ . Find the exact solution set using logarithms and approximate solutions to 4 decimal places.

Assumptions
1. The given equation is $7 e^{2 m-1}-3=11$ .
2. We need to solve for $m$ .
3. The solution should be expressed in terms of common or natural logarithms.
4. An approximate solution should be provided to at least 4 decimal places.

First, we need to isolate the exponential term on one side of the equation. To do this, we will add $3$ to both sides of the equation.
$7 e^{2 m-1} - 3 + 3 = 11 + 3$

Simplify both sides of the equation.
$7 e^{2 m-1} = 14$

Next, we will divide both sides of the equation by $7$ to solve for the exponential term $e^{2 m-1}$ .
$\frac{7 e^{2 m-1}}{7} = \frac{14}{7}$

Simplify both sides of the equation.
$e^{2 m-1} = 2$

Now, we will take the natural logarithm of both sides of the equation to solve for $2m - 1$ .
$\ln(e^{2 m-1}) = \ln(2)$

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the left side of the equation.
$2 m - 1 = \ln(2)$

Next, we will add $1$ to both sides of the equation to solve for $2m$ .
$2 m - 1 + 1 = \ln(2) + 1$

Simplify both sides of the equation.
$2 m = \ln(2) + 1$

Now, we will divide both sides of the equation by $2$ to solve for $m$ .
$m = \frac{\ln(2) + 1}{2}$

This is the exact solution in terms of natural logarithms. We can now find an approximate solution by evaluating the right side of the equation using a calculator.
$m \approx \frac{\ln(2) + 1}{2}$

Using a calculator, we find the approximate value of $\ln(2)$ .
$\ln(2) \approx 0.6931$

Substitute the approximate value of $\ln(2)$ into the equation.
$m \approx \frac{0.6931 + 1}{2}$

Calculate the approximate value of $m$ .
$m \approx \frac{1.6931}{2}$

Simplify the approximation.
$m \approx 0.84655$

Round the approximation to at least 4 decimal places.
$m \approx 0.8466$
The exact solution set is $\left\{\frac{\ln(2) + 1}{2}\right\}$ . $m \approx 0.8466$