Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

for the love of God help me !! I'm desperate for it tomorrow

For The Love Of God Help Me Im Desperate For It Tomorrow class=

Sagot :

konrad509
[tex]D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}[/tex]
superkoopasirf
[tex]\log_2(5x - 2) - \log_2 x - \log_2(x - 1) = 2[/tex]
[tex]\log_2 ( \frac{5x - 2}{x(x-1)} ) = 2[/tex]
[tex]\log_2 ( \frac{5x - 2}{x(x-1)} ) = \log_2 4[/tex]
[tex]\frac{5x - 2}{x(x-1)} = 4[/tex]
[tex]5x - 2 = 4x(x - 1)[/tex]
[tex]5x - 2 = 4x^2 - 4x[/tex]
[tex]0 = 4x^2 - 9x + 2[/tex]
[tex]0 = (4x - 1)(x - 2)[/tex]
[tex]\implies x = 2 \text{ or } x = \frac{1}{4} [/tex]

By putting one quarter into the equation we can see that it cannot work in any way involving real numbers as [tex]\log_2 ( \frac{1}{4} - 1) [/tex] would be log of a negative number, so x = 2.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.