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If log(subscript 10) 3 is approximately .4771, evaluate log (subscript 10) .0003 Give...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted August 15, 2013 at 1:00 AM via web

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If log(subscript 10) 3 is approximately .4771, evaluate log (subscript 10) .0003

Give answer correct to four decimals.

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ishpiro | Teacher | (Level 1) Associate Educator

Posted August 15, 2013 at 1:12 AM (Answer #1)

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The logarithm with the base 10 is usually denoted as log.

log3 = .4771

The number 0.0003 can be written as `3*10^(-4)` , which is a product. The product rule for logarithms states

`log_b xy=log_b x + log_b y`

Thus, log(.0003) can be rewritten as `log(.0003) = log(3*10^(-4)) = log3 + log(10^(-4))`

Log3 is given and `log(10^(-4)) = -4` from the definition of a logarithm.

Therefore log(.0003) = .4771 - 4 = -3.5229

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 15, 2013 at 2:07 AM (Answer #2)

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We know that `log_m(x/y) = log_mx - log_my`

Now, `log_10(3) = 0.4771`

Hence, `log_10(0.0003) = log_10(3/10000) = log_10(3) - log_10(10000)`                                                       `= 0.4771 - 4`

                                                      `=-3.5229` Ans.

                                {Since, `log_10(10000) = log_10(10^4) = 4`

`and log_10(10) = 1` }

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