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Answer 1 · 2017-04-07T07:20:22

$- \infty < x < 1$ $-oo< x<1$

Explanation:

Solve the function for $y$ $y$ : $y = {log}_{a} (a - a^{x})$ $y=log_a(a-a^x)$ .

Since the logarithm can only accept values greater than $0$ $0$ , $a - a^{x} > 0$ $a-a^x>0$ , or $a^{x} < a$ $a^x< a$ . We can take the natural logarithm of both sides to get $x ln (a) < ln (a)$ $xln(a)< ln(a)$ .

Since $a > 1$ $a>1$ , $ln (a) > 0$ $ln(a)>0$ . We can divide both sides by $ln (a)$ $ln(a)$ without worrying about flipping the inequality sign or dividing by $0$ $0$ . Therefore, $x ln (a) < ln (a)$ $xln(a)< ln(a)$ becomes $x < 1$ $x<1$ .

Thus, the domain is $- \infty < x < 1$ $-oo< x<1$ .

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