Certainly! Let's solve the given equation step-by-step:
We are given the equation:
[tex]\[ 3^x + \frac{1}{3^x} = 3 \frac{1}{3} \][/tex]
First, we can represent the right-hand side of the equation as a single fraction:
[tex]\[ 3 + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3} \][/tex]
Rewriting the equation, we get:
[tex]\[ 3^x + \frac{1}{3^x} = \frac{10}{3} \][/tex]
Next, let [tex]\( y = 3^x \)[/tex]. Consequently, [tex]\(\frac{1}{3^x} = \frac{1}{y} \)[/tex]. Substituting these into our equation, the equation becomes:
[tex]\[ y + \frac{1}{y} = \frac{10}{3} \][/tex]
To clear the fraction, multiply both sides of the equation by [tex]\( y \)[/tex]:
[tex]\[ y^2 + 1 = \frac{10}{3}y \][/tex]
Let's rearrange this equation to form a standard quadratic equation:
[tex]\[ 3y^2 + 3 = 10y \][/tex]
Subtract [tex]\( 10y \)[/tex] from both sides to get:
[tex]\[ 3y^2 - 10y + 3 = 0 \][/tex]
Now we have a quadratic equation in the form [tex]\( ay^2 + by + c = 0 \)[/tex] with [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 3 \)[/tex].
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/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 possible solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{10 + 8}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ y = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
Now we have [tex]\( y = 3 \)[/tex] and [tex]\( y = \frac{1}{3} \)[/tex]. Recall that [tex]\( y = 3^x \)[/tex].
For [tex]\( y = 3 \)[/tex]:
[tex]\[ 3^x = 3 \][/tex]
Taking the logarithm base 3 of both sides:
[tex]\[ x = \log_3{3} = 1 \][/tex]
For [tex]\( y = \frac{1}{3} \)[/tex]:
[tex]\[ 3^x = \frac{1}{3} \][/tex]
Taking the logarithm base 3 of both sides:
[tex]\[ x = \log_3{\left(\frac{1}{3}\right)} = -1 \][/tex]
Thus, the solutions to the equation [tex]\(3^x + \frac{1}{3^x} = 3 \frac{1}{3}\)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
