Solving Logarithmic Problems with 0-1 Calculator

[Physicsrapper]

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?

 

[jedishrfu]

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?

 

[Mark44]

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.

 

[HallsofIvy]

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 [itex]0.758856 \times 10^{3}[/itex] and then [itex]log(0.758856)= log(0.758856 \times 10^4)= 4+ log(0.758856)[/itex].

 

[Mark44]

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 [itex]0.758856 \times 10^{3}[/itex] and then [itex]log(0.758856)= log(0.758856 \times 10^4)= 4+ log(0.758856)[/itex].

That last line should be
##log(7588.56)= log(0.758856 \times 10^4)= 4+ log(0.758856)##