Question

1.Simplify log √27 – log8 –log √100
Log 6 - log 5
2.Find the value of x if √x2 – 3x + 6 - √x2- 3x + 3 = 1

Answer 1

1)  Log √27 - Log 8 - Log √100 + log 6 - Log 5

$=Log (3^3)^{\frac{1}{2}}-Log\ 2^3-Log\ 10+Log2*3-Log5\\\\=\frac{3}{2}Log3-3Log2-(Log2+Log5)+(Log 2+log3)-Log5\\\\=\frac{5}{2}Log3-3Log2-2Log5\\\\=Log[ \frac{9\sqrt3}{200} ]$

2.
$\sqrt{x^2-3x+6}-\sqrt{x^2-3x+3}=1\\\\Let\ x^2-3x+3=X\\\\=> \ \ \sqrt{X+3}-\sqrt{X}=1\\=> \ \ \sqrt{X+3}=1+\sqrt{X}\\=> \ \ X+3=1+X+2\sqrt{X}\\=> \ \ 2=2\sqrt{X}\\=>\ X=1\\\\So\ x^2-3x+3=1\\\\x^2-3x+2=0\\\\(x-2)(x-1)=0\\\\x=2\ or\ 1$