If 3^x - 3^{x - 1} = 6,then x^x is equal to | vedclass

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(B) Given equation: $3^x - 3^{x - 1} = 6$We can rewrite the equation as:$3^x - \frac{3^x}{3} = 6$Let $3^x = t$. Substituting this into the equation:$t

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$4$

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(B) Given equation: $3^x - 3^{x - 1} = 6$We can rewrite the equation as:$3^x - \frac{3^x}{3} = 6$Let $3^x = t$. Substituting this into the equation:$t - \frac{t}{3} = 6$Multiply the entire equation by $3$:$3t - t = 18$$2t = 18$$t = 9$Since $t = 3^x$,we have:$3^x = 9 = 3^2$Therefore,$x = 2$.Now,calculate $x^x$:$x^x = 2^2 = 4$.

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

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