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aliya0001 [1]
10 months ago
8

Log(x⁴+3x³) - log(X + 3 ) + log2 - log6 = 2logx . find the value of x

Mathematics
1 answer:
AfilCa [17]10 months ago
7 0

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}

hence

x=0\text{ or x=3}

But x cannot be zero so x=3

So the value of x is 3

h

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