$D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}$

8 0

2 years ago

Read 2 more answers

A number is multiplied by 2, and then 5 is subtracted. The result is 43.

irga5000 [103]

Answer: The answer is 7.

5 0

2 years ago

Rewrite using standard notation : 8.07 x 10^3

sammy [17]

8070, because when it is a positive exponent you have to move the exponent to the right. When it is a negative exponent, you would move it to the left. Hope this helps :)

4 0

2 years ago

The volume of a cube is 441cm3.<br> Work out the length of its side rounded to 1 DP.

Rainbow [258]

Answer:

volume of cube=441cm³

l³=441cm³

$\sqrt[3]{441}$

length=7.6cm

3 0

2 years ago

Prove that : tan^2(theta) - tan^2(alpha) = sec^2(theta) - sec^2(alpha)​

Prove that : tan^2(theta) - tan^2(alpha) = sec^2(theta) - sec^2(alpha)