Solving Logarithmic Problems with 0-1 Calculator

  • Thread starter Thread starter Physicsrapper
  • Start date Start date
  • Tags Tags
    Logarithmic
Click For Summary

Homework Help Overview

The discussion revolves around evaluating logarithmic expressions using a calculator that can only compute logarithms of numbers between 0 and 1. Participants explore how to express various logarithms in a suitable form for this limitation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants suggest rewriting logarithmic expressions by factoring out powers of ten to fit the calculator's constraints. Questions arise regarding the correctness of these transformations and whether additional logarithmic values need to be looked up to complete the evaluations.

Discussion Status

The discussion is ongoing, with participants providing insights into the historical context of logarithm calculations and confirming the validity of the proposed transformations. There is an acknowledgment of the need to look up certain logarithmic values, but no consensus has been reached on the completeness of the solutions presented.

Contextual Notes

Participants reference the limitations of the calculator and the historical practice of using logarithm tables, which only included values for numbers between 0 and 1. This context influences the approach to solving the logarithmic problems.

Physicsrapper
Messages
24
Reaction score
0
Suppose your pocket calculator is damaged: it can only evaluate logarithms of numbers between 0 and 1. Find a way to evaluate the following logarithms with the use of such a calculator.

log2 = log (10*0.2) = log10 + log0.2 = 1 + log0.2

log333 = log(1000 * 0.333) = log10^3 + log0.333 = 3 + log0.333

log1.1 = log(10 * 0.11) = log10 + log0.11 = 1 + log0.11

log7588.56 = log(10 000 * 0.758856) = log10^4 + log0.758856 = 4 + log0.758856

Are these solutions correct?
 
Physics news on Phys.org
They look okay.

Don't you still have to look up the log(0.333) to complete the answer for 3 + log0.333 as an example?
 
jedishrfu said:
They look okay.

Don't you still have to look up the log(0.333) to complete the answer for 3 + log0.333 as an example?
I believe the goal of the exercise is to reduce log expressions to a form for which they can be calculated by the defective calculator, but not to actually do the calculation.
 
Once upon a time- in the years "B.C." (Before Calculators) it was common to look up logarithms in tables- which only gave the logarithms for 0 to 1. To find the logarithm of a number such as 7588.56, yes, you would write it as 0.758856 \times 10^{3} and then log(0.758856)= log(0.758856 \times 10^4)= 4+ log(0.758856).
 
Last edited by a moderator:
HallsofIvy said:
Once upon a time- in the years "B.C." (Before Calculators) it was common to look up logarithms in tables- which only gave the logarithms for 0 to 1. To find the logarithm of a number such as 7588.56, yes, you would write it as 0.758856 \times 10^{3} and then log(0.758856)= log(0.758856 \times 10^4)= 4+ log(0.758856).
That last line should be
##log(7588.56)= log(0.758856 \times 10^4)= 4+ log(0.758856)##
 

Similar threads

Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
17
Views
3K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K