Question-108240

Question Number 108240 by mathdave last updated on 15/Aug/20

Answered by john santu last updated on 15/Aug/20

\frac{J S △}{★}

2^{a + 4} = 6 \Rightarrow 2^{a + 4} = 2.3

\Rightarrow 2^{a + 3} = 3 \dots (i)

3^{2 b + 1} = 3 . 2^{2} \Rightarrow 3^{2 b} = 2^{2}; 3^{b} = 2

t h e n {(2^{a + 3})}^{2 b} = 2^{2}

2^{2 a b + 6 b} = 2^{2} \Rightarrow a b + 3 b = 1

\Rightarrow a b - 1 = - 3 b; a b - 2 = - 3 b - 1

w e o b t a i n 3^{a b - 2} = 3^{- 3 b - 1} = \frac{1}{{(3^{b})}^{3} \times 3}

3^{a b - 2} = \frac{1}{24} .

Answered by 1549442205PVT last updated on 16/Aug/20

2^{a + 4} = 6 \Rightarrow (a + 4) \ln 2 = \ln (2.3) = \ln 2 + \ln 3

\Rightarrow aln 2 = \ln 3 - 3 \ln 2 = \ln \frac{3}{8} \Rightarrow a = \frac{\ln \frac{3}{8}}{\ln 2} (1)

3^{2 b + 1} = 12 \Rightarrow (2 b + 1) \ln 3 = \ln (3.4) \Leftrightarrow

2 bln 3 + \ln 3 = \ln 3 + \ln 4 \Rightarrow 2 bln 3 = \ln 4

\Rightarrow b = \frac{\ln 4}{2 \ln 3} = \frac{2 \ln 2}{2 \ln 3} = \frac{\ln 2}{\ln 3} (2)

From (1) (2) we get ab = \frac{\ln \frac{3}{8}}{\ln 2} \times \frac{\ln 2}{\ln 3}

= \frac{\ln \frac{3}{8}}{\ln 3} = \log_{3} \frac{3}{8}

\Rightarrow ab - 2 = \log_{3} \frac{3}{8} - 2 = \log_{3} \frac{3}{8} - \log_{3} 3^{2}

= \log_{3} (\frac{3}{8} / 9) = \log_{3} (\frac{1}{24}) . Therefore

3^{ab - 2} = 3^{\log_{3} (\frac{1}{24})} = \frac{1}{24} (since a^{\log_{a} N} = N)

Consequently we choose C