Solve the inequality log base 5 of (x - 3 times the square root of x + 32) is less than 2 times log base 5 of (the square root of x + 2), and give the smallest positive integer solution.

Step-by-step solution

1. 1

   Fix the domain. The radical forces $x\ge0$. With $t=\sqrt{x}\ge0$, the first argument is $t^2-3t+32$, whose discriminant is $9-128=-119<0$; since the leading coefficient is positive, it is always positive. The second argument $\sqrt{x}+2\ge2>0$ as well. So the domain is simply $x\ge0$ and no candidate solutions will be lost later.

2. 2

   Pull the coefficient 2 inside the logarithm. Using $k\log_b u=\log_b u^k$:

   $2\log_5(\sqrt{x}+2)=\log_5\left((\sqrt{x}+2)^2\right)$

   Now both sides are a single logarithm with the same base, which is the only form in which arguments can be compared directly.

3. 3

   Drop the logarithms. Because the base $5>1$, $\log_5$ is increasing, so the inequality between logs is equivalent to the same inequality between arguments (the direction is preserved; with a base below $1$ it would flip):

   $x-3\sqrt{x}+32 < (\sqrt{x}+2)^2$

4. 4

   Expand and cancel. The right side is $x+4\sqrt{x}+4$, and the $x$ terms cancel on both sides:

   $-3\sqrt{x}+32 < 4\sqrt{x}+4 \quad\Longrightarrow\quad 28 < 7\sqrt{x} \quad\Longrightarrow\quad \sqrt{x}>4$

5. 5

   Square safely. Both sides are non-negative, so squaring preserves the inequality:

   $x>16$

6. 6

   Read off the smallest integer. The smallest integer strictly greater than $16$ is $x=17$, and it lies in the domain $x\ge0$. Checking: $17-3\sqrt{17}+32\approx36.63$ while $(\sqrt{17}+2)^2\approx37.4924$, so the inequality does hold at $x=17$ but fails at $x=16$ (both sides equal $36$).