Suppose that we know the logarithms of all numbers which are expressed to base 'a' and we are required to find the logarithms of all these numbers to base 'b'. We proceed as follows. Let N be any one of the numbers of which we are required to find the logarithm to base 'b' and the value itself be some 'x'. That is, logbN = x or N = bx. But we already know the value of loga N. Also logaN can be expressed as loga(bx) as N = bx. By rule 5, loga(bx) can be expressed as x.logab or
xlogab = logaN
...............(1)
Since the values of N and b are known, the values of loga N and loga b can be found from the tables. These values when substituted in equation (1) gives us the value of logb N.
In the above equation, what will happen if N = a. Equation (1) will be
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(because logaa = 1)
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or
logba x loga b = 1
Example
Find the values of the following.
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log3 81
We know that 81 = 34. Therefore, log3 34 = 4. log3 = 4.1 = 4
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log3 (9 x 27 x 81)
log3 (9 x 27 x 81) = log39 + log327 + log381
(loga M.N = loga M + logaN)
= log3(32) + log3(33) + log3(34)
= 2.log33 + 3.log33 + 4.log33
= 2 + 3 + 4 = 9
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In this example we apply an extension of rule 5.
(5/4).log33 = (5/4).1 = (5/4)
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log3 (243/81)
log3243 - log381 [loga (M/N) = logaM - loga N]
log335 - log334
5.log33 - 4.log33 = 5 - 4 = 1
(logM (MP) = p.logM M = p)