Question

The sum of all the integers in the domain of the real valued function \[ f(x)=\sqrt{\log_{\frac13}\left(\frac{9x-30}{x-26}\right)} \]

• A. 6
• B. 9
• C. 12
• D. 10

Hint:

For \[ \sqrt{\log_a(f(x))} \] always use \[ \log_a(f(x))\ge0. \] If \[ 0<a<1, \] then \[ \boxed{\log_a t\ge0 \iff 0<t\le1.} \] If \[ a>1, \] then \[ \boxed{\log_a t\ge0 \iff t\ge1.} \]

Answer 1

The Correct Option is A

Step 1: Condition for the square root. Since the logarithm is inside a square root, \[ \log_{\frac13}\left(\frac{9x-30}{x-26}\right)\ge 0. \] As the base of the logarithm is \[ \frac13<1, \] we use the property \[ \log_a t\ge0 \iff 0<t\le1,\qquad (0<a<1). \] Hence, \[ 0< \frac{9x-30}{x-26} \le1. \]

Step 2: Solve the inequality \(\dfrac{9x-30}{x-26}>0\). Factor the numerator: \[ 9x-30=3(3x-10). \] Critical points are \[ x=\frac{10}{3},\qquad x=26. \] Using the sign chart, \[ \frac{9x-30}{x-26}>0 \] for \[ x26. \]

Step 3: Solve the inequality \(\dfrac{9x-30}{x-26}\le1\). \[ \frac{9x-30}{x-26}\le1 \] \[ \frac{9x-30-(x-26)}{x-26}\le0 \] \[ \frac{8x-4}{x-26}\le0 \] \[ \frac{2x-1}{x-26}\le0. \] The critical points are \[ x=\frac12,\qquad x=26. \] Hence, \[ \frac{2x-1}{x-26}\le0 \] for \[ \frac12\le x<26. \]

Step 4: Find the domain. Intersecting the two conditions, \[ \left(x26\right) \cap \left(\frac12\le x<26\right) = \left[\frac12,\frac{10}{3}\right). \] Thus, \[ \boxed{\text{Domain}=\left[\frac12,\frac{10}{3}\right).} \] The integers in the domain are \[ 1,\;2,\;3. \] Their sum is \[ 1+2+3=6. \] Hence, \[ \boxed{6} \] is the correct answer.