Answer :
Let's solve the equation step-by-step to find the value of [tex]\(x\)[/tex]:
Given equation:
[tex]\[ 7\left(3^x\right) + 3^x + 3^x = 9^x \times 27\left(3^x\right) \times 9\left(3^x\right) \][/tex]
First, let's simplify the left side of the equation:
[tex]\[ 7\left(3^x\right) + 3^x + 3^x = 7(3^x) + 2(3^x) \][/tex]
[tex]\[ 7(3^x) + 2(3^x) = (7 + 2)(3^x) \][/tex]
[tex]\[ 9(3^x) \][/tex]
So the left side of the equation is:
[tex]\[ 9(3^x) \][/tex]
Now, let's simplify the right side of the equation. Notice that [tex]\(9^x\)[/tex] and [tex]\(27\)[/tex] can be rewritten using base 3:
[tex]\[ 9 = 3^2 \quad \Rightarrow \quad 9^x = (3^2)^x = 3^{2x} \][/tex]
[tex]\[ 27 = 3^3 \][/tex]
So the right side of the equation becomes:
[tex]\[ 3^{2x} \times 3^3 \times (3^x) \times 9(3^x) \][/tex]
We already know that:
[tex]\[ 9 = 3^2 \quad \Rightarrow \quad 9(3^x) = (3^2)(3^x) = 3^{2+x} \][/tex]
Thus, the entire right side of the equation can be rewritten as:
[tex]\[ 3^{2x} \times 3^3 \times (3^x) \times 3^{2+x} \][/tex]
[tex]\[ = 3^{2x} \times 3^3 \times 3^x \times 3^{2+x} \][/tex]
[tex]\[ = 3^{2x + 3 + x + 2 + x} \][/tex]
[tex]\[ = 3^{2x + 3 + x + 2 + x} \][/tex]
[tex]\[ = 3^{4x + 5} \][/tex]
Therefore, the equation simplifies to:
[tex]\[ 9(3^x) = 3^{4x+5} \][/tex]
Simplifying further, notice that [tex]\(9\)[/tex] can be written as [tex]\(3^2\)[/tex]:
[tex]\[ (3^2)(3^x) = 3^{4x + 5} \][/tex]
[tex]\[ 3^{2 + x} = 3^{4x + 5} \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ 2 + x = 4x + 5 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 2 + x = 4x + 5 \][/tex]
[tex]\[ x - 4x = 5 - 2 \][/tex]
[tex]\[ -3x = 3 \][/tex]
[tex]\[ x = -1 \][/tex]
So the solution to the equation is:
[tex]\[ x = -1 \][/tex]
Given equation:
[tex]\[ 7\left(3^x\right) + 3^x + 3^x = 9^x \times 27\left(3^x\right) \times 9\left(3^x\right) \][/tex]
First, let's simplify the left side of the equation:
[tex]\[ 7\left(3^x\right) + 3^x + 3^x = 7(3^x) + 2(3^x) \][/tex]
[tex]\[ 7(3^x) + 2(3^x) = (7 + 2)(3^x) \][/tex]
[tex]\[ 9(3^x) \][/tex]
So the left side of the equation is:
[tex]\[ 9(3^x) \][/tex]
Now, let's simplify the right side of the equation. Notice that [tex]\(9^x\)[/tex] and [tex]\(27\)[/tex] can be rewritten using base 3:
[tex]\[ 9 = 3^2 \quad \Rightarrow \quad 9^x = (3^2)^x = 3^{2x} \][/tex]
[tex]\[ 27 = 3^3 \][/tex]
So the right side of the equation becomes:
[tex]\[ 3^{2x} \times 3^3 \times (3^x) \times 9(3^x) \][/tex]
We already know that:
[tex]\[ 9 = 3^2 \quad \Rightarrow \quad 9(3^x) = (3^2)(3^x) = 3^{2+x} \][/tex]
Thus, the entire right side of the equation can be rewritten as:
[tex]\[ 3^{2x} \times 3^3 \times (3^x) \times 3^{2+x} \][/tex]
[tex]\[ = 3^{2x} \times 3^3 \times 3^x \times 3^{2+x} \][/tex]
[tex]\[ = 3^{2x + 3 + x + 2 + x} \][/tex]
[tex]\[ = 3^{2x + 3 + x + 2 + x} \][/tex]
[tex]\[ = 3^{4x + 5} \][/tex]
Therefore, the equation simplifies to:
[tex]\[ 9(3^x) = 3^{4x+5} \][/tex]
Simplifying further, notice that [tex]\(9\)[/tex] can be written as [tex]\(3^2\)[/tex]:
[tex]\[ (3^2)(3^x) = 3^{4x + 5} \][/tex]
[tex]\[ 3^{2 + x} = 3^{4x + 5} \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ 2 + x = 4x + 5 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 2 + x = 4x + 5 \][/tex]
[tex]\[ x - 4x = 5 - 2 \][/tex]
[tex]\[ -3x = 3 \][/tex]
[tex]\[ x = -1 \][/tex]
So the solution to the equation is:
[tex]\[ x = -1 \][/tex]
