Question

If $|a^x-2|+|8-a^x|<5$ , then $x\in$____ , where $a\in (1,\infty )$

• A. $\varphi \left(\text{empty set}\right)$
• B. $R$
• C. $(1,\infty )$
• D. $[1,\infty )$

Answer 1

The correct option is A $\varphi \left(\text{empty set}\right)$

Assume $a^x=t$
Then $|t-2|+|8-t|<5$
Minimum value of $|a|+|b|\text{is}|a+b|$
$\Rightarrow min(|t-2|+|8-t|)=|t-2+8-t|=|6|>5$
Hence, no solution exists.