The sum of all the solutions of the equation | vedclass

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Question

(C) Given equation: $(8)^{2x} - 16 \cdot (8)^x + 48 = 0$Let $8^x = t$. Then the equation becomes:$t^2 - 16t + 48 = 0$Factoring the quadratic equation:

Vedclass · Accepted Answer

$1 + \log_8(6)$

Vedclass · Answer

(C) Given equation: $(8)^{2x} - 16 \cdot (8)^x + 48 = 0$Let $8^x = t$. Then the equation becomes:$t^2 - 16t + 48 = 0$Factoring the quadratic equation:$(t - 4)(t - 12) = 0$So,$t = 4$ or $t = 12$.Substituting back $8^x = t$:$8^x = 4 \implies x = \log_8(4)$$8^x = 12 \implies x = \log_8(12)$The sum of the solutions is:$\log_8(4) + \log_8(12) = \log_8(4 	imes 12) = \log_8(48)$Since $48 = 8 	imes 6$,we have:$\log_8(8 	imes 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6)$

The sum of all the solutions of the equation $(8)^{2x} - 16 \cdot (8)^x + 48 = 0$ is:

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