MargGambrelidn
Answered

IDNLearn.com is designed to help you find the answers you need quickly and easily. Our platform is designed to provide quick and accurate answers to any questions you may have.

I need the solution for: log(3x) = log 2 + log(x + 4)

Sagot :

Naǫidn
[tex]\hbox{the domain:} \\ 3x>0 \ \land \ x+4>0 \\ x>0 \ \land \ x>-4 \\ x \in (0,+\infty) \\ \\ \log(3x)=\log 2+\log(x+4) \\ 0=\log 2+ \log(x+4) - \log(3x) \\ 0=\log(\frac{2 \times (x+4)}{3x}) \\ 0=\log(\frac{2x+8}{3x}) \\ 10^0=\frac{2x+8}{3x} \\ 1=\frac{2x+8}{3x} \\ 3x=2x+8 \\ 3x-2x=8 \\ \boxed{x=8}[/tex]
[tex]D:\\3x > 0\ and\ x+4 > 0\\x > 0\ and\ x > -4\\therefore\ D:x\in(-4;\ \infty)\\------------------\\log(3x)=log2+log(x+4)\ \ \ |use\ loga+logb=log(a\cdot b)\\\\log(3x)=log[2(x+4)]\iff3x=2(x+4)\\\\3x=2x+8\ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\\boxed{x=8\in D}[/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.