Sprawdzenie prostego równania logarytmicznego
: 30 sie 2011, o 15:06
\(\displaystyle{ \log _{ \sqrt{3} } \left\{ \log _{3} \left[ \log _{2} \left( x+2 \right) \right] \right\} = 0}\)
DZIEDZINA:
\(\displaystyle{ x + 2 > 0 \\
x > -2 \\
\\
\log _{2} (x + 2 ) > 0 \\
\log _{2} (x + 2 ) > \log _{2} 1 \\
x + 2 > 1 \\
x > -1 \\
\\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] > 0 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] > \log _{3} 1 \\
\log_{2} \left( x + 2 \right) > 1 \\
\log_{2} \left( x + 2 \right) > \log_{2} 2 \\
x + 2 > 2 \\
x > 0 \\
\\
D = \left( 0 ; + \infty \right)}\)
ROZWIĄZANIE:
\(\displaystyle{ \log _{ \sqrt{3} } \left\{ \log _{3} \left[ \log _{2} \left( x+2 \right) \right] \right\} = \log _{ \sqrt{3} } 1 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] = 1 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] = \log _{3} 3 \\
\log _{2} \left( x+2 \right) = 3 \\
\log _{2} \left( x+2 \right) = \log _{2} 8 \\
x+2 = 8 \\
x = 6}\)
Wszystko dobrze?
DZIEDZINA:
\(\displaystyle{ x + 2 > 0 \\
x > -2 \\
\\
\log _{2} (x + 2 ) > 0 \\
\log _{2} (x + 2 ) > \log _{2} 1 \\
x + 2 > 1 \\
x > -1 \\
\\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] > 0 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] > \log _{3} 1 \\
\log_{2} \left( x + 2 \right) > 1 \\
\log_{2} \left( x + 2 \right) > \log_{2} 2 \\
x + 2 > 2 \\
x > 0 \\
\\
D = \left( 0 ; + \infty \right)}\)
ROZWIĄZANIE:
\(\displaystyle{ \log _{ \sqrt{3} } \left\{ \log _{3} \left[ \log _{2} \left( x+2 \right) \right] \right\} = \log _{ \sqrt{3} } 1 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] = 1 \\
\log _{3} \left[ \log_{2} \left( x + 2 \right) \right] = \log _{3} 3 \\
\log _{2} \left( x+2 \right) = 3 \\
\log _{2} \left( x+2 \right) = \log _{2} 8 \\
x+2 = 8 \\
x = 6}\)
Wszystko dobrze?