Let f(x) = 2^{10} cdot x + 1 and g(x) = 3^{10 | vedclass

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(D) Given $f(x) = 2^{10}x + 1$ and $g(x) = 3^{10}x - 1$. We are given $(f \circ g)(x) = x$. Substituting $g(x)$ into $f(x)$,we get $f(g(x)) = 2^{10}(3

Vedclass · Accepted Answer

$\frac{1 - 2^{-10}}{3^{10} - 2^{-10}}$

Vedclass · Answer

(D) Given $f(x) = 2^{10}x + 1$ and $g(x) = 3^{10}x - 1$. We are given $(f \circ g)(x) = x$. Substituting $g(x)$ into $f(x)$,we get $f(g(x)) = 2^{10}(3^{10}x - 1) + 1 = x$. Expanding the expression: $2^{10} \cdot 3^{10}x - 2^{10} + 1 = x$. Since $2^{10} \cdot 3^{10} = (2 \cdot 3)^{10} = 6^{10}$,we have $6^{10}x - x = 2^{10} - 1$. Factoring out $x$: $x(6^{10} - 1) = 2^{10} - 1$. Thus,$x = \frac{2^{10} - 1}{6^{10} - 1}$. To match the options,divide the numerator and denominator by $6^{10}$: $x = \frac{\frac{2^{10}}{6^{10}} - \frac{1}{6^{10}}}{1 - \frac{1}{6^{10}}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$. Alternatively,divide the numerator and denominator by $2^{10} \cdot 3^{10} = 6^{10}$ or manipulate the expression: $x = \frac{2^{10} - 1}{6^{10} - 1} = \frac{2^{10}(1 - 2^{-10})}{6^{10}(1 - 6^{-10})} = \frac{3^{-10}(1 - 2^{-10})}{1 - 6^{-10}}$. Re-evaluating the expression $\frac{2^{10} - 1}{6^{10} - 1}$: Divide numerator and denominator by $6^{10}$: $\frac{2^{10}/6^{10} - 1/6^{10}}{1 - 1/6^{10}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$. Checking option $D$: $\frac{1 - 2^{-10}}{3^{10} - 2^{-10}} = \frac{(2^{10}-1)/2^{10}}{(3^{10} \cdot 2^{10} - 1)/2^{10}} = \frac{2^{10}-1}{6^{10}-1}$. Thus,option $D$ is correct.

Let $f(x) = 2^{10} \cdot x + 1$ and $g(x) = 3^{10} \cdot x - 1$. If $(f \circ g)(x) = x$,then $x$ is equal to

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