Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

please help me I'm desperate.

Please Help Me Im Desperate class=

Sagot :

konrad509
[tex]D:4x-1>0 \wedge x+2 >0 \wedge x>0\\ D:4x>1 \wedge x>-2 \wedge x>0\\ D:x>\frac{1}{4} \wedge x>0\\ D:x>\frac{1}{4}\\\\ \log(4x-1)-\log(x+2)=\log x\\ \log\frac{4x-1}{x+2}=\log x\\ \frac{4x-1}{x+2}=x\\ x(x+2)=4x-1\\ x^2+2x=4x-1\\ x^2-2x+1=0\\ (x-1)^2=0\\ \boxed{x=1} [/tex]
superkoopasirf
[tex]\log (4x - 1) - \log (x + 2) = \log x[/tex]
[tex]\log ( \frac{4x - 1}{x + 2} ) = \log x[/tex]
[tex] \frac{4x - 1}{x + 2} = x[/tex]
[tex]4x - 1 = x(x + 2)[/tex]
[tex]4x - 1 = x^2 + 2x[/tex]
[tex]0 = x^2 - 2x + 1[/tex]
[tex]0 = (x - 1)^2[/tex]
⇒ [tex]x = 1[/tex]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.