What is the value of log100*log99*…*log1
Answers
What is the solution of Log(1/2) +log(2/3) + log(3/4) +…+log(8/9) +log (9/10)?
Slight correction, you are not after a solution but the value of an expression.
To evaluate your expression, I just need to use three simple points about logarithms, namely:
[math]\log(1) = 0[/math]
[math]\log (a/b) = \log(a) - \log(b)[/math]
[math]\log_n(n) = 1[/math]
Using point 2 to rewrite your expression, we have:
[math][\log(1) - \log(2)] + [\log(2) - \log(3)] + ... + [\log(8) - \log(9)] + [\log(9) - \log(10)][/math]
[math]= \log(1) - \log(10)[/math]
Using point 2, this simplifies to: [math]-\log(10)[/math]
The actual answer depends on what base logarithm you are using. I will assume that you are using common logs (aka decimal logs or base-10 logs), in which case, from point 3, the expression reduces to [math]-1[/math]
[tex][/tex]