Bài 6: Bất phương trình mũ và logarit

\(9^x-\left(2m+2\right)3^x-2m-3>0\)

\(\Leftrightarrow9^x-2.3^x-3-2m\left(3^x+1\right)>0\)

\(\Leftrightarrow\left(3^x+1\right)\left(3^x-3\right)-2m\left(3^x+1\right)>0\)

\(\Leftrightarrow\left(3^x+1\right)\left(3^x-2m-3\right)>0\)

\(\Leftrightarrow3^x-2m-3>0\)

\(\Leftrightarrow2m< 3^x-3\Rightarrow m< \min\limits_R\left(f\left(x\right)\right)\)

Với \(f\left(x\right)=3^x-3\)

Do \(3^x>0\Rightarrow f\left(x\right)>-3\Rightarrow2m\le-3\Rightarrow m\le-\frac{3}{2}\)

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: \(x>1\)

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

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

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

\(\Leftrightarrow x>2\)

ĐKXĐ: \(\left\{{}\begin{matrix}x^2+4x-5>0\\x+7>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-7< x< -5\\x>1\end{matrix}\right.\)

Khi đó BPT tương đương:

\(log_2\left(x^2+4x-5\right)>2log_{2^{-1}}\left(\frac{1}{x+7}\right)\)

\(\Leftrightarrow log_2\left(x^2+4x-5\right)>log_2\left(x+7\right)^2\)

\(\Leftrightarrow x^2+4x-5>x^2+14x+49\)

\(\Leftrightarrow10x< -54\Rightarrow x< -\frac{27}{5}\)

Kết hợp ĐKXĐ \(\Rightarrow-\frac{27}{5}< x< -5\Rightarrow a=-\frac{27}{5};b=-5\)

\(\Rightarrow...\)

\(\Leftrightarrow5\left(\frac{9^x}{25^x}\right)+2\left(\frac{15^x}{25^x}\right)-3\ge0\)

\(\Leftrightarrow5\left(\frac{3}{5}\right)^x+2\left(\frac{3}{5}\right)^x-3\ge0\)

\(\Leftrightarrow\left[5\left(\frac{3}{5}\right)^x-3\right]\left[\left(\frac{3}{5}\right)^x+1\right]\ge0\)

\(\Leftrightarrow5\left(\frac{3}{5}\right)^x-3\ge0\)

\(\Leftrightarrow\left(\frac{3}{5}\right)^x\ge\frac{3}{5}\)

\(\Rightarrow x\ge1\)

Đáp án B

ĐKXĐ: \(x>0\)

\(\Leftrightarrow log_2^2\left(2x\right)+log_2\left(2x\right)-log_28-9< 0\)

\(\Leftrightarrow log_2^2\left(2x\right)+log_2\left(2x\right)-12< 0\)

\(\Leftrightarrow\left(log_2\left(2x\right)+4\right)\left(log_2\left(2x\right)-3\right)< 0\)

\(\Leftrightarrow-4< log_2\left(2x\right)< 3\)

\(\Leftrightarrow\frac{1}{16}< 2x< 8\Leftrightarrow\frac{1}{32}< x< 4\)