(Poor Person's Logarithm) For this question, we will be given a number x and a base b and will find the smallest positive integer y such that by > x. For example, if x = 10 and b = 2, we would have y = 4 since 2³ = 8 < 10 < 2ª = 16. In order to solve this via loops, DO NOT use the built in logarithm functions. poor_log Function: Input variables: • a scalar representing x • a scalar representing the base b Output variables: • a scalar representing the output y described above A possible sample case is: > y = poor_log(8, 2) y = 3 >> y = poor_log(10, 2) y = 4 >> y = poor_log(16, 2) y = 4
(Poor Person's Logarithm) For this question, we will be given a number x and a base b and will find the smallest positive integer y such that by > x. For example, if x = 10 and b = 2, we would have y = 4 since 2³ = 8 < 10 < 2ª = 16. In order to solve this via loops, DO NOT use the built in logarithm functions. poor_log Function: Input variables: • a scalar representing x • a scalar representing the base b Output variables: • a scalar representing the output y described above A possible sample case is: > y = poor_log(8, 2) y = 3 >> y = poor_log(10, 2) y = 4 >> y = poor_log(16, 2) y = 4
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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make c+++ code
![(Poor Person’s Logarithm) For this question, we will be given a number \( x \) and a base \( b \) and will find the smallest positive integer \( y \) such that
\[ b^y \geq x. \]
For example, if \( x = 10 \) and \( b = 2 \), we would have \( y = 4 \) since \( 2^3 = 8 < 10 \leq 2^4 = 16 \).
In order to solve this via loops, DO NOT use the built-in logarithm functions.
**poor_log Function:**
- **Input variables:**
- a scalar representing \( x \)
- a scalar representing the base \( b \)
- **Output variables:**
- a scalar representing the output \( y \) described above
A possible sample case is:
```
>> y = poor_log(8, 2)
y = 3
>> y = poor_log(10, 2)
y = 4
>> y = poor_log(16, 2)
y = 4
```](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36268169-052e-482b-acb5-c5dcae700f3f%2F7994c39b-01e0-45fe-b1fc-7700a2293644%2F9z8phvv_processed.png&w=3840&q=75)
Transcribed Image Text:(Poor Person’s Logarithm) For this question, we will be given a number \( x \) and a base \( b \) and will find the smallest positive integer \( y \) such that
\[ b^y \geq x. \]
For example, if \( x = 10 \) and \( b = 2 \), we would have \( y = 4 \) since \( 2^3 = 8 < 10 \leq 2^4 = 16 \).
In order to solve this via loops, DO NOT use the built-in logarithm functions.
**poor_log Function:**
- **Input variables:**
- a scalar representing \( x \)
- a scalar representing the base \( b \)
- **Output variables:**
- a scalar representing the output \( y \) described above
A possible sample case is:
```
>> y = poor_log(8, 2)
y = 3
>> y = poor_log(10, 2)
y = 4
>> y = poor_log(16, 2)
y = 4
```
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