Question

If $\frac{x^{a^2}}{x^{b^2}}=x^{16},x>1,$ and $a+b=2$, what is the value of $a-b$?

• A. 8
• B. 14
• C. 16
• D. 18

Answer 1

The correct option is A 8

Given, $\frac{x^{a^2}}{x^{b^2}}=x^{16}$

Also given, $x>1$ and $a+b=2$ ..(1)

$\Rightarrow x^{a^2-b^2}=x^{16}$

$\Rightarrow a^2-b^2=16$ ...(2) (When the non-zero base is the same, the index also must be the same)

Dividing equations (1) and (2),

$\therefore \frac{a^2-b^2}{a+b}=a-b=\frac{16}{2}=8$