IDNLearn.com offers a reliable platform for finding accurate and timely answers. Get step-by-step guidance for all your technical questions from our dedicated community members.
Sagot :
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].
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].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
