Question

The number of positive real roots of the equation
\[ 3^{x+1}+3^{-x+1}=10 \] is

• A. \(3\)
• B. \(2\)
• C. \(1\)
• D. Infinitely many

Hint:

For equations involving \(a^x\) and \(a^{-x}\), substitute \(t=a^x\). Then \(a^{-x}=\frac{1}{t}\), which converts the equation into a quadratic equation.

Answer 1

The Correct Option is C

Step 1: Rewrite the equation.
Given equation is
\[ 3^{x+1}+3^{-x+1}=10 \]
Now,
\[ 3^{x+1}=3\cdot 3^x \]
and
\[ 3^{-x+1}=3\cdot 3^{-x} \]
Therefore,
\[ 3\cdot 3^x+3\cdot 3^{-x}=10 \]

Step 2: Substitute \(t=3^x\).
Let
\[ t=3^x \]
Since \(3^x\gt 0\), we have
\[ t\gt 0 \]
Also,
\[ 3^{-x}=\frac{1}{3^x}=\frac{1}{t} \]
So the equation becomes
\[ 3t+\frac{3}{t}=10 \]

Step 3: Convert into quadratic equation.
Multiplying by \(t\),
\[ 3t^2+3=10t \]
\[ 3t^2-10t+3=0 \]

Step 4: Solve the quadratic equation.
\[ 3t^2-10t+3=0 \]
Factorizing,
\[ 3t^2-9t-t+3=0 \]
\[ 3t(t-3)-1(t-3)=0 \]
\[ (t-3)(3t-1)=0 \]
Thus,
\[ t=3 \] or
\[ t=\frac13 \]

Step 5: Find values of \(x\).
Since \(t=3^x\),
For \(t=3\),
\[ 3^x=3 \]
\[ x=1 \]
For \(t=\frac13\),
\[ 3^x=\frac13=3^{-1} \]
\[ x=-1 \]

Step 6: Count positive real roots.
The roots are
\[ x=1,\;-1 \]
Among these, only \(x=1\) is positive.
Therefore, the number of positive real roots is
\[ 1 \]

Step 7: Final conclusion.
Hence,
\[ \boxed{1} \]