how to find cube root of a number using logarithms???

Asked by  | 29th Sep, 2008, 09:58: AM

Expert Answer:

Let x = 3.7510-12

x1/=  [3.7510-12] 1/3

Taking logs on both sides , we have

logx1/= log [3.7510-12] 1/3

i.e 1/3 log x =1/3 log 3.75 - 12/3 log 10

=1/3 log x =1/3 log 3.75 - 4    (since log 10 = 1)

 log x =3[1/3 log 3.75 - 4 ]

log x = log 3.75 -12

x = antilog [ log 3.75 -12]

I assume you can  find logs and antilogs and substitute

Answered by  | 29th Sep, 2008, 11:32: AM

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