Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
[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]
[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]
[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]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.