The number of positive real roots of the equa | vedclass

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(C) Given the equation $3^{x+1}+3^{-x+1}=10$. This can be rewritten as $3(3^x) + \frac{3}{3^x} = 10$. Let $y = 3^x$. Then the equation becomes $3y + \

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$1$

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(C) Given the equation $3^{x+1}+3^{-x+1}=10$. This can be rewritten as $3(3^x) + \frac{3}{3^x} = 10$. Let $y = 3^x$. Then the equation becomes $3y + \frac{3}{y} = 10$. Multiplying by $y$,we get $3y^2 - 10y + 3 = 0$. Factoring the quadratic equation: $3y^2 - 9y - y + 3 = 0 \Rightarrow 3y(y-3) - 1(y-3) = 0$. So,$(3y-1)(y-3) = 0$,which gives $y = 3$ or $y = \frac{1}{3}$. Substituting back $y = 3^x$: Case $1$: $3^x = 3^1 \Rightarrow x = 1$. Case $2$: $3^x = 3^{-1} \Rightarrow x = -1$. The positive real root is $x = 1$. Therefore,there is only $1$ positive real root.

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

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