Rewrite the equation to simplify the terms: 18 ( 6 ) 2 x + 9 ( 6 ) x − 2 = 0 .
Substitute y = 6 x into the equation: 18 y 2 + 9 y − 2 = 0 .
Solve the quadratic equation for y : y = − 3 2 and y = 6 1 .
Since 0"> y = 6 x > 0 , we have y = 6 1 , so x = − 1 .
The final answer is − 1 .
Explanation
Problem Analysis We are given the equation ( 2 2 x + 1 ) ( 3 2 x + 2 ) + 2 x ( 3 x + 2 ) − 2 = 0 . Our goal is to find the value(s) of x that satisfy this equation.
Simplifying the Equation First, let's rewrite the equation to simplify the terms:
( 2 2 x × 2 ) ( 3 2 x × 3 2 ) + 2 x ( 3 x × 3 2 ) − 2 = 0
2 × 9 × ( 2 2 x 3 2 x ) + 9 × ( 2 x 3 x ) − 2 = 0
18 ( 2 × 3 ) 2 x + 9 ( 2 × 3 ) x − 2 = 0
18 ( 6 ) 2 x + 9 ( 6 ) x − 2 = 0
Substitution Let y = ( 6 ) x . Substituting y into the equation, we get a quadratic equation:
18 y 2 + 9 y − 2 = 0
Solving the Quadratic Equation Now, we solve the quadratic equation 18 y 2 + 9 y − 2 = 0 for y . The roots of this quadratic equation are:
y = 2 ( 18 ) − 9 ± 9 2 − 4 ( 18 ) ( − 2 ) = 36 − 9 ± 81 + 144 = 36 − 9 ± 225 = 36 − 9 ± 15
So, y 1 = 36 − 9 − 15 = 36 − 24 = − 3 2 and y 2 = 36 − 9 + 15 = 36 6 = 6 1 .
Finding x Since y = 6 x , y must be positive. Therefore, y = − 3 2 is not a valid solution. We only consider y = 6 1 .
Now, we solve 6 x = 6 1 for x .
6 x = 6 − 1
Thus, x = − 1 .
Final Answer Therefore, the solution to the equation is x = − 1 .
Examples
Exponential equations are used in various real-world applications, such as modeling population growth, radioactive decay, and compound interest. For instance, if you invest money in a bank account with compound interest, the amount of money you have after a certain period can be modeled using an exponential equation. Similarly, in radioactive decay, the amount of a radioactive substance remaining after a certain time can be modeled using an exponential equation. Understanding how to solve exponential equations is crucial for making predictions and decisions in these scenarios.
To solve the equation ( 2 2 x + 1 ) ( 3 2 x + 2 ) + 2 x ( 3 x + 2 ) − 2 = 0 , we simplified it to a quadratic form and found the valid solution, which is x = − 1 .
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