Find the solution of the exponential equation [tex]$11 e^x-14=3$[/tex]
The exact solution, in terms of the natural logarithm is: [tex]$x=$[/tex]
The approximate solution, accurate to 4 decimal places is: [tex]$x=$[/tex]
Answer (2)
Isolate the exponential term by adding 14 to both sides and dividing by 11: e x = 11 17 .
Take the natural logarithm of both sides: x = ln ( 11 17 ) .
Calculate the approximate value of the natural logarithm to 4 decimal places: x ≈ 0.4353 .
The exact solution is x = ln ( 11 17 ) and the approximate solution is 0.4353 .
Explanation
Problem Analysis We are given the exponential equation 11 e x − 14 = 3 . Our goal is to find both the exact solution in terms of the natural logarithm and the approximate solution accurate to 4 decimal places.
Isolating the Exponential Term First, we isolate the exponential term. Add 14 to both sides of the equation: 11 e x = 3 + 14
Simplifying Simplify the right side: 11 e x = 17
Isolating e^x Next, divide both sides by 11 to isolate e x :
e x = 11 17
Taking the Natural Logarithm Now, take the natural logarithm of both sides of the equation: ln ( e x ) = ln ( 11 17 )
Finding the Exact Solution Using the property that ln ( e x ) = x , we find the exact solution: x = ln ( 11 17 )
Finding the Approximate Solution To find the approximate solution, we calculate the value of ln ( 11 17 ) accurate to 4 decimal places. The result of this calculation is approximately 0.4353.
Final Answer Therefore, the exact solution is x = ln ( 11 17 ) , and the approximate solution is x ≈ 0.4353 .
Examples
Exponential equations are used in various fields such as finance, physics, and engineering. For example, they can model population growth, radioactive decay, and compound interest. Understanding how to solve exponential equations allows us to predict future values, determine decay rates, and calculate investment returns. In finance, we can use exponential equations to calculate the time it takes for an investment to double at a certain interest rate. In physics, we can use them to determine the half-life of a radioactive substance.
The exact solution to the equation 11 e x − 14 = 3 is x = ln ( 11 17 ) , and the approximate solution is x ≈ 0.4353 .
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