Question

1) log_0,5(2x-3) > -1

2) log₈(x-1) + log₈(x+2) > \(\frac{2}{3}\)

Answer 1

1.

ĐKXĐ: \(x>\frac{3}{2}\)

\(\Leftrightarrow-log_2\left(2x-3\right)>-1\)

\(\Leftrightarrow log_2\left(2x-3\right)< 1\)

\(\Leftrightarrow2x-3< 2\)

\(\Leftrightarrow x< \frac{5}{2}\)

Vậy \(\frac{3}{2}< x< \frac{5}{2}\)

2.

ĐKXĐ: \(x>1\)

\(log_8\left(x-1\right)\left(x+2\right)>\frac{2}{3}\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)>8^{\frac{2}{3}}=4\)

\(\Leftrightarrow x^2+x-6>0\)

\(\Leftrightarrow x>2\)