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Before calculators were around, why was it advantageous to use logarithms to calculate roots?

Before calculators were around, why was it advantageous to use logarithms to calculate roots?

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Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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Hi Bella,
This is an interesting question, since logarithms have been around for much longer than calculators!
Before calculators were around, people used tables to calculate anything that went beyond the basic arithmetic operations of addition, subtraction, multiplication, and division. Square roots and cube roots were known since ancient times. There were tables for square roots, cube roots, fourth roots etc for sure, but they allowed for only a limited number of values to be computed.
Then, in the 16th century, an English mathematician called John Napier discovered logarithms. As you may know, logarithms have one amazing property: they turn products into sums, as in log(a*b)=log(a)+log(b). That means exponents become factors, as in log(a2)=2 log(a), and log(√a)=1/2 log(a) (remember that taking a square root is the same as raising that number to the power 1/2). Napier tabulated thousands of logarithms and developed a simple device, called a slide rule, that lets you compute any logarithm with 3-digit or higher accuracy. Many mathematicians after Napier refined his tables to so many values that they fill entire books. So if you wanted to compute, say, √2 you would use the fact that log(√2)= (1/2) log(2). You'd look up the value of log(2) in the table, get 0.3010 (usually they tabulated the decadic, or base-10, logarithms), divide it by 2, get 0.1505, then look up which log would give you 0.1505, and it would be 1.414, which is √2 rounded to 4 significant digits. It's really neat!


Nice reminder of history and Napier.
Oh the day's of a slide rule (some scales are logarithmic) to multiply or do "advanced" math.
Thanks, Edward! My classroom actually still had a huge wooden model of a slide rule hanging on the wall -- I found that fascinating!
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
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Hi Bella;
This is the way I understand the situation.
The invention of calculators is a non-issue.  Logarithms were designed to facilitate graphing in a circumstance which the quantity of one variable has exponential changes, and that of the other variable has few changes.
For example, let's say that Any-town of Any-state in the USA was evaluated each decade by the United States Census Bureau to have population statistics of...
1900, 1,000
1910, 10,000
1920, 100,000
We can graph this by characterizing the x-axis as the decade, and the y-axis as population.  It would require a long piece-of-paper.  Or, we can use logarithms to characterize the statistics as...
1900, log101000=3
1910, log1010000=4
1920, log10100000=5
That would require a standard piece-of-paper the size of 8.5 x 11-inches.
In simpler terms, it is not about speed of calculations, but rather size of paper.
Edward B. | All Levels Math, Computer Science & Computer Programming, SAT MathAll Levels Math, Computer Science & Comp...
4.0 4.0 (1 lesson ratings) (1)
Logarithm is the mathematical inverse of exponentiation.
Exponentiation is the mathematical inverse of logarithm.
Best with an example: SQRT(9) which is 3.
Sorry about not being able to mathematically typeset this, but this is a simple editor.
x = SQRT(9)
x = 9 ^ (1/2)   since logs are mathematical inverses of exponentiation
                        take log of both sides
log(x) = log (9^(1/2))
log(x) = (1/2) log (9)   now before calculators, you would look up log(9) in a log table
log (x) ~ (1/2) (0.9542)
log (x) ~ 0.47712   since we want x, we do the anti-log [log^(-1) or 10^x], so back to the table
                             to find what value yields a log of 0.47712

log^(-1) log (x) ~ log^(-1) 0.47712
x = 3
The above is for the square root.   But this works for all real numbers.
What about the cube root of 10 squared? 
Well that's the cube root of 100, which is between 4 and 5, but what about an exact answer.
Trial and error is endless.
x = (10^2) ^ (1/3)
x = 10 ^ (2/3)
log(x) = log [10 ^ (2/3)]
log(x) = (2/3) log(10)
log(x) = (2/3) 1
log(x) ~ 0.6666666...
log^(-1) log (x) ~ log^(-1) 0.6666666...
x ~ 4.64159