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March 21, 2020

Challenge Problem 9: Clever Logarithm Problem

Simplify the following product: (log_2 3)(log_3 4)(log_4 5)…(log_127 128)

For u >0 and a base b that is positive but not equal to 1, we can use the change of base formula to express the value in terms of the natural logarithm:

log_b u = (ln u)/(ln b)

Using the formula for the product we get:

(log_2 3)(log_3 4)(log_4 5)…(log_127 128)
= [(ln 3)/(ln 2)][(ln 4)/(ln 3)][(ln 5)/(ln 4)]…[(ln 128)/(ln 127)]

This is a telescoping product: the numerator of term k cancels with the denominator of the subsequent k + 1 term. Only the last numerator and the first denominator survive, so we get the result:

= (ln 128)/(ln 2)
= log_2 128
= log_2 2^7
= 7

(The second step used the change of base formula log_2 128 = (ln 128)/(ln 2))

And that’s it! The entire product miraculously simplifies to lucky number 7.


Adapted from Reddit HomeworkHelp
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Yêu Tiếng Anh


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