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

hence

$x=0\text{ or x=3}$

But x cannot be zero so x=3

So the value of x is 3

7 0

10 months ago

Neil has 3 partially full cans of white paint. They contain 1/3 of a gallon and 1/5 and 1/2 of a gallon of paint. About how much

allsm [11]

Answer:

1¹/₃₀

Step-by-step explanation:

The 3 gallons were partially full

They contain 1/3 of a gallon and 1/5 and 1/2 of a gallon of paint.

the first is 1/3 of the gallon

the second is 1/5 of galloon

the third is 1/2 of gallon adding them together

1/3 +1/5 +1/2 = (10+6+15)/30 = 31/30 = 1¹/₃₀

3 0

3 years ago