Solving Logarithmic Problems with 0-1 Calculator

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The discussion centers on evaluating logarithmic expressions using a damaged calculator that only computes logarithms of numbers between 0 and 1. Participants confirm that the proposed solutions for various logarithms, such as log2 and log7588.56, are correct in their reduced forms. However, there is a clarification that while the expressions can be simplified, the actual logarithm values for numbers like 0.333 still need to be looked up for complete answers. The exercise aims to demonstrate how to express logarithms in a way that accommodates the calculator's limitations, reminiscent of historical methods using logarithm tables. The conversation emphasizes understanding the reduction process rather than performing the calculations directly.
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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?
 
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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)##
 

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