Solve the inequality log base 5 of (x - 3*sqrt(x) + 32) < 2 * log base 5 of (sqrt(x) + 2), and state the least positive integer solution. Options: 17, 25, 16, 20.

Step-by-step solution

1. 1

   Establish the domain first, since logarithms restrict it. The presence of $\sqrt{x}$ already forces $x\geq 0$. For the left logarithm we need $x-3\sqrt{x}+32>0$; substituting $t=\sqrt{x}\geq 0$ turns this into $t^{2}-3t+32>0$, whose discriminant is

   $D=(-3)^{2}-4(1)(32)=9-128=-119<0$

   With a negative discriminant and a positive leading coefficient, $t^{2}-3t+32$ is positive for every $t$, so this imposes nothing new. The right logarithm needs $\sqrt{x}+2>0$, which is automatic. Hence the domain is exactly $x\geq 0$.

2. 2

   Fold the coefficient $2$ into the logarithm. Using $k\log_{b}u=\log_{b}u^{k}$, valid because $\sqrt{x}+2>0$:

   $2\log_{5}\left(\sqrt{x}+2\right)=\log_{5}\left(\left(\sqrt{x}+2\right)^{2}\right)$

   Now both sides are single base-$5$ logarithms.

3. 3

   Drop the logarithms in the correct direction. Since the base $5>1$, $\log_{5}$ is strictly increasing, so the inequality between logarithms is equivalent to the same inequality between arguments — with no sign flip:

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

   If the base were between $0$ and $1$, this step would reverse the sign.

4. 4

   Expand and watch the $x$ terms cancel. Since $\left(\sqrt{x}+2\right)^{2}=x+4\sqrt{x}+4$,

   $x-3\sqrt{x}+32<x+4\sqrt{x}+4$

   Subtracting $x$ from both sides removes the quadratic-in-$\sqrt{x}$ part entirely and leaves a linear inequality in $\sqrt{x}$:

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

5. 5

   Square safely and intersect with the domain. Both sides of $\sqrt{x}>4$ are nonnegative, so squaring preserves the inequality:

   $x>16$

   This already lies inside the domain $x\geq 0$, so the solution set is $(16,+\infty)$.

6. 6

   Pick the least positive integer and test it. The smallest integer strictly greater than $16$ is $17$, which is option 1. Direct substitution confirms the strictness of the boundary: at $x=16$ the two sides are $\log_{5}(36)$ and $\log_{5}(36)$, equal and therefore not a solution, while at $x=17$ the left side is $\log_{5}(36.63)$ and the right is $\log_{5}(37.49)$, so the inequality holds. Testing $x=0,1,4,9,16$ all fail and $16.5,17,25$ all pass, matching $x>16$.