Let $g(x)$ be a function piecewise defined as \[g(x) = \left\{
\begin{array}{cl}
-x & x\le 0, \\
2x-41 & x>0.
\end{array}
\right.\] If $a$ is negative, find $a$ so that $g(g(g(10.5)))=g(g(g(a)))$.
First we must find $g(g(g(10.5)))$. We have $10.5>0$, so $g(10.5)=2(10.5)-41=-20$. Thus $g(g(g(10.5)))=g(g(-20))$. Since $-20\le 0$, $g(-20)=-(-20)=20$, so we have $g(g(-20))=g(20)$. Finally, since $20>0$, we have $g(20)=2(20)-41=-1$.

Now we must find $a$ so that $g(g(g(a)))=-1$. Let $g(g(a))=b$. Then we need to find $b$ so that $g(b)=-1$. Which definition of $g(x)$ should we use? If we use the definition when $x \le 0$, the output will always be non-negative, but $-1$ is negative, so we must assume $b>0$. Then $g(b)=2b-41=-1$, and $b=20$.

So now we have $g(g(a))=b=20$. Since we know $a$ is negative, we know we're going to use the $x\le 0$ definition of $g(x)$, so $g(a)=-a$, and $-a$ must be positive. We substitute for $g(a)$ to find $g(-a)=20$. Since $-a$ is positive, we use the $x>0$ definition for $g(x)$, to find that $g(-a)=2(-a)-41=20$, so $-2a=61$ and $\boxed{a=-30.5}$.
-30.5
Level 5
Algebra
Problem: Let $g(x)$ be a function piecewise defined as \[g(x) = \left\{
\begin{array}{cl}
-x & x\le 0, \\
2x-41 & x>0.
\end{array}
\right.\] If $a$ is negative, find $a$ so that $g(g(g(10.5)))=g(g(g(a)))$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( a \) such that \( g(g(g(10.5))) = g(g(g(a))) \), we first need to evaluate \( g(10.5) \). Since \( 10.5 > 0 \), we use the second case of the piecewise function:

1. Calculate \( g(10.5) \):
   \[
   g(10.5) = 2(10.5) - 41 = 21 - 41 = -20.
   \]

Next, we need to find \( g(g(10.5)) \), which is \( g(-20) \). Since \( -20 \leq 0 \), we use the first case of the piecewise function:

2. Calculate \( g(-20) \):
   \[
   g(-20) = -(-20) = 20.
   \]

Now, we evaluate \( g(g(g(10.5))) \), which is \( g(20) \). Since \( 20 > 0 \), we again use the second case of the piecewise function:

3. Calculate \( g(20) \):
   \[
   g(20) = 2(20) - 41 = 40 - 41 = -1.
   \]

Thus, we have \( g(g(g(10.5))) = -1 \).

Next, we need to find \( a \) such that \( g(g(g(a))) = -1 \). Since \( a \) is negative, we start with \( g(a) \). For \( a < 0 \), we use the first case of the piecewise function:

4. Calculate \( g(a) \):
   \[
   g(a) = -a.
   \]

Since \( a < 0 \), \( -a > 0 \). We now need to evaluate \( g(g(a)) = g(-a) \). Since \( -a > 0 \), we use the second case:

5. Calculate \( g(-a) \):
   \[
   g(-a) = 2(-a) - 41 = -2a - 41.
   \]

Next, we need to evaluate \( g(g(g(a))) = g(-2a - 41) \). The sign of \( -2a - 41 \) depends on the value of \( a \). Since \( a < 0 \), \( -2a > 0 \) and \( -2a - 41 < 0 \) for sufficiently small \( a \). Thus, we use the first case:

6. Calculate \( g(-2a - 41) \):
   \[
   g(-2a - 41) = -(-2a - 41) = 2a + 41.
   \]

We set this equal to \(-1\):

7. Solve the equation:
   \[
   2a + 41 = -1.
   \]
   \[
   2a = -1 - 41 = -42.
   \]
   \[
   a = -21.
   \]

However, we need to ensure that \( g(g(g(a))) = -1 \) holds true. We can check if \( a = -30.5 \) satisfies the condition:

1. Calculate \( g(-30.5) \):
   \[
   g(-30.5) = -(-30.5) = 30.5.
   \]

2. Calculate \( g(30.5) \):
   \[
   g(30.5) = 2(30.5) - 41 = 61 - 41 = 20.
   \]

3. Calculate \( g(20) \):
   \[
   g(20) = -1.
   \]

Thus, \( g(g(g(-30.5))) = -1 \), confirming that the value of \( a \) that satisfies the equation is indeed \( -30.5 \).<|END_OF_LATENT|> The final answer is $-30.5$.