$(2^{2 x+1})(3^{2 x+2})+2^x(3^{x+2})-2=0
Answer (2)
Rewrite the given exponential equation in terms of 6 x .
Substitute y = 6 x to obtain a quadratic equation 18 y 2 + 9 y − 2 = 0 .
Solve the quadratic equation for y to get y = 6 1 and y = − 3 2 .
Solve for x using 6 x = y , which gives x = − 1 as the only real solution since 6 x cannot be negative. 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.
Rewriting the Equation Let's rewrite the equation to make it easier to work with. We can rewrite the terms as follows:
2 2 x + 1 = 2 ⋅ 2 2 x = 2 ⋅ ( 2 2 ) x = 2 ⋅ 4 x 3 2 x + 2 = 3 2 ⋅ 3 2 x = 9 ⋅ ( 3 2 ) x = 9 ⋅ 9 x 2 x = 2 x 3 x + 2 = 3 2 ⋅ 3 x = 9 ⋅ 3 x
Substituting these back into the original equation, we get:
( 2 ⋅ 4 x ) ( 9 ⋅ 9 x ) + 2 x ( 9 ⋅ 3 x ) − 2 = 0 18 ⋅ 4 x ⋅ 9 x + 9 ⋅ 2 x ⋅ 3 x − 2 = 0 18 ⋅ ( 4 ⋅ 9 ) x + 9 ⋅ ( 2 ⋅ 3 ) x − 2 = 0 18 ⋅ ( 36 ) x + 9 ⋅ ( 6 ) x − 2 = 0
Substitution Now, let's make a substitution to simplify the equation further. Let y = 6 x . Then, y 2 = ( 6 x ) 2 = 6 2 x = ( 6 2 ) x = 3 6 x . Substituting this into our equation, we get:
18 y 2 + 9 y − 2 = 0
Solving the Quadratic Equation We now have a quadratic equation in terms of y . We can solve this using the quadratic formula:
y = 2 a − b ± b 2 − 4 a c
In our equation, a = 18 , b = 9 , and c = − 2 . Plugging these values into the quadratic formula, we get:
y = 2 ( 18 ) − 9 ± 9 2 − 4 ( 18 ) ( − 2 ) y = 36 − 9 ± 81 + 144 y = 36 − 9 ± 225 y = 36 − 9 ± 15
So, we have two possible values for y :
y 1 = 36 − 9 + 15 = 36 6 = 6 1 y 2 = 36 − 9 − 15 = 36 − 24 = − 3 2
Solving for x Now, we need to solve for x . Recall that we made the substitution y = 6 x . So, we have two equations to solve:
6 x = 6 1 6 x = − 3 2
For the first equation, we can rewrite 6 1 as 6 − 1 . So, we have:
6 x = 6 − 1 x = − 1
For the second equation, 6 x = − 3 2 , there is no real solution since 6 x is always positive for any real value of x .
Final Answer Therefore, the only real solution to the original equation is x = − 1 .
Examples
Exponential equations like this appear in various fields, such as modeling population growth, radioactive decay, and compound interest. For instance, if you're analyzing the spread of a virus, the number of infected individuals might be modeled by an exponential function. Solving equations involving these functions helps predict when the number of cases will reach a certain threshold, which is crucial for public health planning. Similarly, in finance, understanding exponential growth is essential for making informed investment decisions.
The equation ( 2 2 x + 1 ) ( 3 2 x + 2 ) + 2 x ( 3 x + 2 ) − 2 = 0 simplifies to a quadratic form when rewritten. The only real solution found is x = − 1 .
;
