Let's solve the equation step by step.
Given equation:
[tex]\[ 3^x + \frac{1}{3^x} = 3 + \frac{1}{3} \][/tex]
First, let's simplify the right-hand side (RHS):
[tex]\[ 3 + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3} \][/tex]
So, the equation becomes:
[tex]\[ 3^x + \frac{1}{3^x} = \frac{10}{3} \][/tex]
Next, introduce a substitution: let [tex]\( y = 3^x \)[/tex]. Hence, [tex]\(\frac{1}{3^x} = \frac{1}{y} \)[/tex].
Substituting this into the equation, we get:
[tex]\[ y + \frac{1}{y} = \frac{10}{3} \][/tex]
Multiply both sides of the equation by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ y^2 + 1 = \frac{10}{3} y \][/tex]
Rearrange this equation into the standard quadratic form:
[tex]\[ 3 y^2 - 10 y + 3 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 3 \)[/tex].
[tex]\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{100 - 36}}{6} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{64}}{6} \][/tex]
[tex]\[ y = \frac{10 \pm 8}{6} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{10 + 8}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ y_2 = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
So, we have:
[tex]\[ y_1 = 3 \][/tex]
[tex]\[ y_2 = \frac{1}{3} \][/tex]
Recall that [tex]\( y = 3^x \)[/tex], hence:
[tex]\[ 3^x = 3 \implies x = \log_3(3) = 1 \][/tex]
[tex]\[ 3^x = \frac{1}{3} \implies x = \log_3\left(\frac{1}{3}\right) = -\log_3(3) = -1 \][/tex]
Thus, the solutions to the given equation are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = -1 \][/tex]
