3 years ago

Find the sum of the finite geometric series -3+6-12+24+-48+96-192+384

Mathematics

1 answer:

hammer [34]3 years ago

3 0

255 (hope this helped)

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AfilCa [17]

The given equation is

$\begin{gathered} \log (x^4+3x^3)-\log (x+3)+\log 2-\log 6=2\log x \\ \log (\frac{x^4+3x^{3^{}}}{x+3})+\log \frac{2}{6}=\log x^2 \\ \log \frac{x^3(x+3)}{x+3}+\log \frac{1}{3}=\log x^2 \\ \log x^3+\log \frac{1}{3}=\log x^2 \\ \log \frac{x^3}{3}=\log x^2 \\ \frac{x^3}{3}=x^2 \\ x^3-3x^2=0 \\ x^2(x-3)=0 \end{gathered}$

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So the value of x is 3

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