Question

How many real numbers satisfy the equation \[{3^{2x + 2}} - {3^{x + 3}} - {3^x} + 3 = 0\] ?

Answer 1

Hint: According to the question, we need to first make the equation simple by splitting the powers. Then, we will assign a variable to a term and solve accordingly. We will factorize the equation and solve for the variable and get the answer.

Complete step-by-step answer:
The given equation is:
 \[{3^{2x + 2}} - {3^{x + 3}} - {3^x} + 3 = 0\]
We can rewrite some terms here. First, we will try to split all the powers here. This will make it easy to solve:
 \[ \Rightarrow {3^{2x}} \times {3^2} - {3^x} \times {3^3} - {3^x} + 3 = 0\]
When we solve this, we get:
 \[ \Rightarrow {3^{2x}} \times 9 - {3^x} \times 27 - {3^x} + 3 = 0\]
We can see here that \[ - {3^x}\] is common here, so we take it outside and put rest of the terms in bracket, and we get:
 \[ \Rightarrow {3^{2x}} \times 9 - {3^x}(27 + 1) + 3 = 0\]
When we solve the bracket, we get:
 \[ \Rightarrow {3^{2x}} \times 9 - {3^x} \times 28 + 3 = 0\]
Now, we will assign a variable to \[{3^x}\] . So, let us assume that:
 \[{3^x} = a\]
Now, when we put the value of \[{3^x}\] as \[a\] in the given equation, then we get:
 \[ \Rightarrow {a^2} \times 9 - 28 \times a + 3 = 0\]
 \[ \Rightarrow 9{a^2} - 28a + 3 = 0\]
Now, we will split \[ - 28a\] into \[ - 27a - a\] , and we get:
 \[ \Rightarrow 9{a^2} - 27a - a + 3 = 0\]
 Now, we will factorize the equation. We will keep the common terms outside the bracket, and the uncommon terms in the bracket, and we get:
 \[ \Rightarrow 9a(a - 3) - 1(a - 3) = 0\]
Now, we will solve the equation, and we get:
 \[ \Rightarrow (a - 3)(9a - 1) = 0\]
Now, we will solve for \[a\] .
When \[a - 3 = 0\] , we get that:
 \[ \Rightarrow a = 3\]
When \[9a - 1 = 0\] , we get that:
 \[ \Rightarrow 9a = 1\]
 \[ \Rightarrow a = \dfrac{1}{9}\]
Now, we have two values for \[a\] which is \[3\] and \[\dfrac{1}{9}\] .
We know that \[{3^x} = a\] . So, we get that the values for \[{3^x}\] are \[3\] and \[\dfrac{1}{9}\] . This tells us that:
 \[{3^x} = 3\,and\,{3^x} = \dfrac{1}{9}\]
Now, we will solve for \[{3^x} = 3\] . When we rewrite this, we get that:
 \[{3^x} = {3^1}\]
We can see that there are same bases on both the sides of the equation, so we will make an equation of the powers, and we get that:
 \[x = 1\]
Similarly, we will solve for \[{3^x} = \dfrac{1}{9}\] . We need to rewrite the equation in such a form, so that there same bases on both the side of the equation, and we get:
 \[{3^x} = \dfrac{1}{{{3^2}}}\]
 \[ \Rightarrow {3^x} = {3^{ - 2}}\]
We can see that there are same bases on both the sides of the equation, so we will make an equation of the powers, and we get that:
 \[ \Rightarrow x = - 2\]
So, the correct answer is “x = 1 AND x = - 2”.

Note: The equation in the given question is a polynomial equation. When a function is in the form of \[f(x) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_n}{x^n}\] , then it is called a polynomial function, and when this function is made equal to \[0\] , it is called a polynomial equation.