Answer :
Sure, let's solve the equation step by step:
Given the equation:
[tex]\[ \frac{3^{2x} + 1}{3^x} = \frac{82}{9} \][/tex]
1. Simplify the left side of the equation:
Notice that the numerator on the left side can be rewritten to simplify the expression:
[tex]\[ \frac{3^{2x} + 1}{3^x} \][/tex]
We know that [tex]\(3^{2x} = (3^x)^2\)[/tex]. So we can rewrite the equation as:
[tex]\[ \frac{(3^x)^2 + 1}{3^x} \][/tex]
We can then separate the terms in the numerator:
[tex]\[ \frac{(3^x)^2}{3^x} + \frac{1}{3^x} = 3^x + \frac{1}{3^x} \][/tex]
2. Rewrite the equation:
Now the original equation becomes:
[tex]\[ 3^x + \frac{1}{3^x} = \frac{82}{9} \][/tex]
3. Introduce a substitution:
Let [tex]\(u = 3^x\)[/tex]. Thus, [tex]\(\frac{1}{3^x} = \frac{1}{u}\)[/tex].
The equation now looks like:
[tex]\[ u + \frac{1}{u} = \frac{82}{9} \][/tex]
4. Multiply through by [tex]\(u\)[/tex]:
To clear the fraction, multiply both sides by [tex]\(u\)[/tex]:
[tex]\[ u^2 + 1 = \frac{82}{9} u \][/tex]
5. Form a quadratic equation:
Rearrange the terms to form a standard quadratic equation:
[tex]\[ 9u^2 - 82u + 9 = 0 \][/tex]
6. Solve the quadratic equation:
To find the roots of the quadratic equation [tex]\(9u^2 - 82u + 9 = 0\)[/tex], we can use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 9\)[/tex], [tex]\(b = -82\)[/tex], and [tex]\(c = 9\)[/tex]. Plugging in these values:
[tex]\[ u = \frac{82 \pm \sqrt{(-82)^2 - 4 \times 9 \times 9}}{2 \times 9} \][/tex]
[tex]\[ u = \frac{82 \pm \sqrt{6724 - 324}}{18} \][/tex]
[tex]\[ u = \frac{82 \pm \sqrt{6400}}{18} \][/tex]
[tex]\[ u = \frac{82 \pm 80}{18} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u = \frac{82 + 80}{18} = \frac{162}{18} = 9 \][/tex]
and
[tex]\[ u = \frac{82 - 80}{18} = \frac{2}{18} = \frac{1}{9} \][/tex]
7. Back-substitute [tex]\(u = 3^x\)[/tex]:
Recall that [tex]\(u = 3^x\)[/tex]. So, we have two equations:
[tex]\[ 3^x = 9 \quad \text{and} \quad 3^x = \frac{1}{9} \][/tex]
Solve for [tex]\(x\)[/tex]:
- For [tex]\(3^x = 9\)[/tex]:
[tex]\[ 9 = 3^2 \implies 3^x = 3^2 \implies x = 2 \][/tex]
- For [tex]\(3^x = \frac{1}{9}\)[/tex]:
[tex]\[ \frac{1}{9} = 3^{-2} \implies 3^x = 3^{-2} \implies x = -2 \][/tex]
8. Conclusion:
The solutions to the equation [tex]\(\frac{3^{2x} + 1}{3^x} = \frac{82}{9}\)[/tex] are:
[tex]\[ x = -2 \quad \text{and} \quad x = 2 \][/tex]
Given the equation:
[tex]\[ \frac{3^{2x} + 1}{3^x} = \frac{82}{9} \][/tex]
1. Simplify the left side of the equation:
Notice that the numerator on the left side can be rewritten to simplify the expression:
[tex]\[ \frac{3^{2x} + 1}{3^x} \][/tex]
We know that [tex]\(3^{2x} = (3^x)^2\)[/tex]. So we can rewrite the equation as:
[tex]\[ \frac{(3^x)^2 + 1}{3^x} \][/tex]
We can then separate the terms in the numerator:
[tex]\[ \frac{(3^x)^2}{3^x} + \frac{1}{3^x} = 3^x + \frac{1}{3^x} \][/tex]
2. Rewrite the equation:
Now the original equation becomes:
[tex]\[ 3^x + \frac{1}{3^x} = \frac{82}{9} \][/tex]
3. Introduce a substitution:
Let [tex]\(u = 3^x\)[/tex]. Thus, [tex]\(\frac{1}{3^x} = \frac{1}{u}\)[/tex].
The equation now looks like:
[tex]\[ u + \frac{1}{u} = \frac{82}{9} \][/tex]
4. Multiply through by [tex]\(u\)[/tex]:
To clear the fraction, multiply both sides by [tex]\(u\)[/tex]:
[tex]\[ u^2 + 1 = \frac{82}{9} u \][/tex]
5. Form a quadratic equation:
Rearrange the terms to form a standard quadratic equation:
[tex]\[ 9u^2 - 82u + 9 = 0 \][/tex]
6. Solve the quadratic equation:
To find the roots of the quadratic equation [tex]\(9u^2 - 82u + 9 = 0\)[/tex], we can use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 9\)[/tex], [tex]\(b = -82\)[/tex], and [tex]\(c = 9\)[/tex]. Plugging in these values:
[tex]\[ u = \frac{82 \pm \sqrt{(-82)^2 - 4 \times 9 \times 9}}{2 \times 9} \][/tex]
[tex]\[ u = \frac{82 \pm \sqrt{6724 - 324}}{18} \][/tex]
[tex]\[ u = \frac{82 \pm \sqrt{6400}}{18} \][/tex]
[tex]\[ u = \frac{82 \pm 80}{18} \][/tex]
This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[ u = \frac{82 + 80}{18} = \frac{162}{18} = 9 \][/tex]
and
[tex]\[ u = \frac{82 - 80}{18} = \frac{2}{18} = \frac{1}{9} \][/tex]
7. Back-substitute [tex]\(u = 3^x\)[/tex]:
Recall that [tex]\(u = 3^x\)[/tex]. So, we have two equations:
[tex]\[ 3^x = 9 \quad \text{and} \quad 3^x = \frac{1}{9} \][/tex]
Solve for [tex]\(x\)[/tex]:
- For [tex]\(3^x = 9\)[/tex]:
[tex]\[ 9 = 3^2 \implies 3^x = 3^2 \implies x = 2 \][/tex]
- For [tex]\(3^x = \frac{1}{9}\)[/tex]:
[tex]\[ \frac{1}{9} = 3^{-2} \implies 3^x = 3^{-2} \implies x = -2 \][/tex]
8. Conclusion:
The solutions to the equation [tex]\(\frac{3^{2x} + 1}{3^x} = \frac{82}{9}\)[/tex] are:
[tex]\[ x = -2 \quad \text{and} \quad x = 2 \][/tex]