Question

If log2=0.3010,log3=0.4771,log7=0.8450,find the value of log84

Answer 1

Answer:

log(84)

=log(2x3x2x7)

=log(2) +log(2) ++log(3) +log(7)

0.301+0.301+0.477+0.845

1.926(approx)

Answer 2

The value of log 84 would be 1.9241

Step-by-step explanation:

Given expression,

log 84

$=\log (12\times 7)$

$=\log (12)+\log 7$    ( ∵ log(mn) = log m + log n )

$=\log (4\times 3) +\log 7$

$=\log 4 +\log 3 +\log 7$

$=\log 2^2 + \log 3 +\log 7$

$=2\log 2 + \log 3 +\log 7$  ( ∵ $\log m^n = n\log m$ )

We have,

log 2 = 0.3010,log 3=0.4771,log 7=0.8450

By substituting the values,

We get,

log 84 = 2 × 0.3010 + 0.4771 + 0.8450 = 0.602 + 1.3221 = 1.9241,

#Learn more:

If log2 =0.30103 & log3 = 0.47712 then the value of log7 is?

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