Problem 28 - Contemporary Precalculus

Question

Solve the equation for \(x\) by first making an appropriate substitution. $$6 e^{2 x}+7 e^{x}=10$$

Step-by-Step Solution

Answer

Question: Solve the exponential equation and find the value of x: $$6 e^{2 x}+7 e^{x}-10=0$$ Answer: The solution to the given exponential equation is $$x = \ln{\frac{1}{2}}$$.

Step 1: Make the substitution

Substitute \(y = e^x\) in the original equation: $$6 e^{2 x}+7 e^{x}-10=0 \implies 6y^2+7y-10=0$$ Now, the equation is converted into a quadratic equation.

Step 2: Solve the quadratic equation

Solve the quadratic equation in terms of \(y\): $$6y^2 + 7y - 10 = 0$$ By finding the factors, we can write the equation as: $$(2y-1)(3y+10)=0$$ Now, we find the respective values of \(y\): $$2y-1=0 \implies y=\frac{1}{2}$$ $$3y+10=0 \implies y=-\frac{10}{3}$$ So, we got two possible values for \(y\): \(y_1 = \frac{1}{2}\) and \(y_2 = -\frac{10}{3}\).

Step 3: Find the values of \(x\)

Now, we need to use the substitution \(y = e^x\) to find the corresponding values for \(x\). For each possible value of \(y\), we find \(x\) as follows: $$y_1 = \frac{1}{2} \implies e^x = \frac{1}{2} \implies x = \ln{\frac{1}{2}}$$ $$y_2 = -\frac{10}{3}$$ Since \(e^x\) can never be equal to a negative value, \(y_2\) is not a valid solution. Therefore, the only solution is: $$x = \ln{\frac{1}{2}}$$

Key Concepts

Exponential Equations

Exponential equations are equations in which variables appear as exponents. These types of equations are common in various fields such as physics, finance, and biology because they describe natural phenomena such as population growth and radioactive decay. In the given exercise, we have an equation in the form of \(6e^{2x} + 7e^x = 10\). Here, \(e^x\) is the exponential part where \(e\) is an important mathematical constant approximately equal to 2.71828.
This type of equation often looks complex due to its exponential nature, but with the right techniques, we can simplify and solve them efficiently. One common approach is to transform the exponential equation into a quadratic form through substitution, making it easier to work with standard algebraic methods.
By understanding how exponential equations function and how they can be manipulated, tackling such problems becomes more manageable. This understanding is key to solving real-world problems that model exponential growth or decay but require a calculation of the precise variable such as time or rate.

Substitution Method

The substitution method is a powerful algebraic technique used to simplify complex equations. It involves replacing a part of the equation with a single variable, making it resemble a more standard form that is easier to solve. In our case, we used substitution to transform an exponential equation into a quadratic equation, which is much simpler to handle.
In this exercise, we replaced \(e^x\) with a new variable \(y\). This allowed us to rewrite the original exponential equation \(6e^{2x} + 7e^x - 10 = 0\) as a quadratic equation: \(6y^2 + 7y - 10 = 0\). Through this substitution, we can now use familiar methods like factoring to solve the equation.

• Identify the part of the equation to substitute (here, \(e^x\)).
• Choose a new variable for substitution (we used \(y\)).
• Rewrite the entire equation using the substitution. This usually transforms it into a more manageable form such as a linear or quadratic equation.

The substitution method is valuable when you need to streamline equations and make complex mathematical relationships easier to solve.

Solving for x

Solving for \(x\) often involves isolating \(x\) on one side of the equation. In our exercise, after performing the substitution and solving the quadratic equation, we returned to solving for \(x\) using our substitution variable.
After finding the values for \(y\), which were \(y_1 = \frac{1}{2}\) and \(y_2 = -\frac{10}{3}\), we then used these results to find the corresponding values for \(x\). Only the positive value of \(y\) was valid because \(e^x\) (an exponential function) cannot equal a negative number.
For \(y_1 = \frac{1}{2}\), we substitute back to find \(x\): \[e^x = \frac{1}{2}\] Solving this gives \(x = \ln{\frac{1}{2}}\), and using properties of logarithms, this can be simplified further if needed.

• Substitute values back to solve for \(x\).
• Use logarithmic identities where necessary to simplify expressions.
• Always check if the solutions make sense in the context of the original equation.

Thus, solving for \(x\) accurately relies on understanding the features of the original function—and using techniques like substitution and logarithms to isolate and determine the value of \(x\).