Question

Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice,

Answer 1

The values of N, log N and ln N are all worked out below.

We have the number N as 8.99

hence;

1) lnN = ln(8.99) = 2.196

log N = log (8.99) = 0.954

2) If we have ln N as 3.949

N = e^3.949 = 51.88

So;

log N = log (51.88) = 1.715

3) log N = 0.134

N = Antilog (0.134 ) = 1.36

ln N = ln (1.36) = 0.307

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