Question

If $1,lo{g}_9(3^{1-x}+2)lo{g}_3({4.3}^x-1)$ are in A.P., then x equals

• A. $lo{g}_{3}4$
• B. $1+lo{g}_{3}4$
• C. $1-lo{g}_{3}4$
• D. $lo{g}_{4}3$

Answer 1

The correct option is C $1-lo{g}_{3}4$

$1,lo{g}_9(3^{1-x}+2),lo{g}_3({4.3}^x-1)areinA.P.\Rightarrow 2lo{g}_9(3^{1-x}+2)=1+lo{g}_3({4.3}^x-1)lo{g}_3(3^{1-x}+2)=lo{g}_{3}3+lo{g}_3({4.3}^x-1)lo{g}_3(3^{1-x}+2)=lo{g}_3\left[3\right({4.3}^x-1\left)\right]3^{1-x}+2=3({4.3}^x-1)(Put{3}^x=t)\frac{3}{t}+2=12t-3or12t^2-5t-3=0;$
Hence $t=-\frac{1}{3},\frac{3}{4}\Rightarrow 3^x=\frac{3}{4}\to x=lo{g}_3\left(\frac{3}{4}\right)orx=lo{g}_{3}3-lo{g}_{3}4\Rightarrow x=1-lo{g}_{3}4$