solve ASAP ---### Evaluation of Postfix Expressions Using a Stack#### Problem StatementEvaluate the following postfix expression using a stack (show detailed steps):\[ 5\ y\ 1\ +\ \log\ \times\ a\ 5\ \div\ - \]#### Note:- The characters **'y'** and **'a'** are considered as variables.- The symbols **'log', '+', '-', '×', '÷'** are considered as operators.---### Detailed Steps1. **Initial Stack State:** - **Empty Stack**2. **Reading from Left to Right:**\[\begin{array}{|l|c|}\hline\text{Expression Element} & \text{Stack State} \\ \hline\ 5 \ & [5] \\ \hline\ y \ & [5, y] \\ \hline\ 1 \ & [5, y, 1] \\ \hline\ + \ & [5, y + 1] \\ \hline\ \log \ & [5, \log(y + 1)] \\ \hline\ \times \ & [5 \times \log(y + 1)] \\ \hline\ a \ & [5 \times \log(y + 1), a] \\ \hline\ 5 \ & [5 \times \log(y + 1), a, 5] \\ \hline\ \div \ & [5 \times \log(y + 1), a \div 5] \\ \hline\ - \ & [5 \times \log(y + 1) - (a \div 5)] \\ \hline\end{array}\]3. **Final Stack State:** - **[5 × log(y + 1) - (a ÷ 5)]**The final result of the evaluation process is **\[5 \times \log(y + 1) - (a \div 5)\]**.---### Explanation of Operators- **+ :** Addition- **- :** Subtraction- **× :** Multiplication- **÷ :** Division- **log :** Logarithm---

Rules for postfix evaluation: Read expression from left to right Push the element in the stack if…

Evaluate the following Postfix expression using a stack (show detailed steps) 5 y 1 + log x a 5 Note: '. The characters 'y' & 'a' are considered as variables. Finally, 'log, +,-, x,+' are considered as operators.

Evaluate the following Postfix expression using a stack (show detailed steps) 5 y 1 + log x a 5 Note: '. The characters 'y' & 'a' are considered as variables. Finally, 'log, +,-, x,+' are considered as operators.

Computer Networking: A Top-Down Approach (7th Edition)

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ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

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### Evaluation of Postfix Expressions Using a Stack

#### Problem Statement
Evaluate the following postfix expression using a stack (show detailed steps):

\[ 5\ y\ 1\ +\ \log\ \times\ a\ 5\ \div\ - \]

#### Note:
- The characters **'y'** and **'a'** are considered as variables.
- The symbols **'log', '+', '-', '×', '÷'** are considered as operators.

---

### Detailed Steps

1. **Initial Stack State:**
- **Empty Stack**

2. **Reading from Left to Right:**

\[
\begin{array}{|l|c|}
\hline
\text{Expression Element} & \text{Stack State} \\ \hline
\ 5 \ & [5] \\ \hline
\ y \ & [5, y] \\ \hline
\ 1 \ & [5, y, 1] \\ \hline
\ + \ & [5, y + 1] \\ \hline
\ \log \ & [5, \log(y + 1)] \\ \hline
\ \times \ & [5 \times \log(y + 1)] \\ \hline
\ a \ & [5 \times \log(y + 1), a] \\ \hline
\ 5 \ & [5 \times \log(y + 1), a, 5] \\ \hline
\ \div \ & [5 \times \log(y + 1), a \div 5] \\ \hline
\ - \ & [5 \times \log(y + 1) - (a \div 5)] \\ \hline
\end{array}
\]

3. **Final Stack State:**
- **[5 × log(y + 1) - (a ÷ 5)]**

The final result of the evaluation process is **\[5 \times \log(y + 1) - (a \div 5)\]**.

---

### Explanation of Operators
- **+ :** Addition
- **- :** Subtraction
- **× :** Multiplication
- **÷ :** Division
- **log :** Logarithm

---

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