# Bài 75 trang 127 SGK giải tích 12 nâng cao

##### Hướng dẫn giải

a) Điều kiện: $$x > 0$$

Ta có: $$lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}\left( {{3^{x + 1}} - 3} \right) = 12$$

\eqalign{ & \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}3\left( {{3^x} - 1} \right) = 12 \cr & \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right)\left[ {1 + lo{g_3}\left( {{3^x} - 1} \right)} \right] = 12 \cr}

$$\Leftrightarrow \log _3^2\left( {{3^x} - 1} \right) + lo{g_3}\left( {{3^x} - 1} \right) - 12 = 0$$

\eqalign{ & \Leftrightarrow \left[ \matrix{ lo{g_3}\left( {{3^x} - 1} \right) = - 4 \hfill \cr lo{g_3}\left( {{3^x} - 1} \right) = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {3^x} - 1 = {1 \over {81}} \hfill \cr {3^x} - 1 = {3^3} = 27 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ {3^x} = {{82} \over {81}} \hfill \cr {3^x} = 28 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\log _3}{{82} \over {81}} \hfill \cr x = {\log _3}28 \hfill \cr} \right. \cr}

Vậy $$S = \left\{ {{{\log }_3}28;{{\log }_3}82 - 4} \right\}$$

b) Điều kiện: $$x > 1$$; $$x \ne 2$$

Ta có: $${\log _{x - 1}}4 = {1 \over {{{\log }_4}\left( {x - 1} \right)}} = {2 \over {{{\log }_2}\left( {x - 1} \right)}}$$. Đặt $$t = {\log _2}\left( {x - 1} \right)$$

Ta có phương trình:

\eqalign{ & {2 \over t} = 1 + t \Leftrightarrow {t^2} + t - 2 = 0 \cr & \Leftrightarrow \left[ \matrix{ t = 1 \hfill \cr t = - 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {\log _2}\left( {x - 1} \right) = 1 \hfill \cr {\log _2}\left( {x - 1} \right) = - 2 \hfill \cr} \right.\left[ \matrix{ x = 3 \hfill \cr x = {5 \over 4} \hfill \cr} \right. \cr}

Vậy $$S = \left\{ {3;{5 \over 4}} \right\}$$

c) Điều kiện: $${\log _2}\left( { - x} \right) \ge 0 \Leftrightarrow - x \ge 1 \Leftrightarrow x \le - 1$$

$$5\sqrt {{{\log }_2}\left( { - x} \right)} = {\log _2}\sqrt {{x^2}} \Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)} = {\log _2}\left( { - x} \right)$$

$$\Leftrightarrow 5\sqrt t = t$$ với $$t = {\log _2}\left( { - x} \right) \ge 0$$

\eqalign{ & \Leftrightarrow 25t = {t^2} \Leftrightarrow \left[ \matrix{ t = 0 \hfill \cr t = 25 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {\log _2}\left( { - x} \right) = 0 \hfill \cr lo{g_2}\left( { - x} \right) = 25 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = - 1 \hfill \cr x = - {2^{25}} \hfill \cr} \right. \cr}

Vậy $$S = \left\{ { - 1; - {2^{25}}} \right\}$$

d) Điều kiện: $$x > 0$$

Ta có: $$\sqrt x = \sqrt {{4^{{{\log }_4}x}}} = {2^{{{\log }_4}x}}$$

Do đó $${3^{{1 \over 2} + {{\log }_4}x}} + {3^{{{\log }_4}x - {1 \over 2}}} = \sqrt x \Leftrightarrow \left( {\sqrt 3 + {1 \over {\sqrt 3 }}} \right){3^{{{\log }_4}x}} = {2^{{{\log }_4}x}}$$

\eqalign{ & \Leftrightarrow {4 \over {\sqrt 3 }} = {\left( {{2 \over 3}} \right)^{{{\log }_4}x}} \Leftrightarrow {\log _4}x = {\log _{{2 \over 3}}}{4 \over {\sqrt 3 }} \cr & \Leftrightarrow x = {4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}} \cr}

Vậy $$S = \left\{ {{4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}}} \right\}$$