Question

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

Answer 1

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 (2)/(6)=\log x^2 \\ \log (x^3(x+3))/(x+3)+\log (1)/(3)=\log x^2 \\ \log x^3+\log (1)/(3)=\log x^2 \\ \log (x^3)/(3)=\log x^2 \\ (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