Logarithms
Logarithmic inequality; base condition
MMTS_Full_Test_22
Grade 12
Question:
Least positive integral value of $a$ for which $\log_{(x+1)/x}(a^2-3a+3)>0$ for all $x>0$
Step-by-Step Solution
Key Concept: For $x>0$: base $(x+1)/x=1+1/x>1$. So $\log_b(A)>0\Leftrightarrow A>1$. Need $a^2-3a+3>1$ for all $x>0$ (independent of $x$): $a^2-3a+2>0\Rightarrow a<1$ or $a>2$.
Least positive integral $a=3$.
Correct Answer: (C) 3