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Why natural log has base e?
Because when the system of logarithms with the base 'e' was defined and tabulated, it was entitled with the identifying label of "Natural Logarithms". ---------------------------------- My improvement: The natural log base is e (a numerical constant of about 2.718). It is chosen as a log base since there is a mathematical series (a "string" of mathematical numerical terms to be summed) for calculating a logarithm (ie. exponent of the base) of a number, which has a base of e. Series for calculating logarithms with bases other than e have basically not been developed.
What is the value of log 50?
Very simple: it is 1.6989700043 to be exact. You can test this because log50 means we assume the natural log (base 10), if you test 10 to the exponent of 1.6989700043 you should render 50 as your result :D
What is the natural log of x?
It means the logarithm to the base e. The number "e" is approximately 2.71828... In other words, if you ask, for instance, "what's the natural logarithm of 100", that's equivalent to asking "to what number must I raise 'e', to get the answer 100". The solution of the equation e^x = 100 in this example.
How do you convert natural log into binary log?
log2x = log x / log 2 On the right side, you can use logarithm in any base (calculators usually provide base-10 and base-e), just be sure to use the same base in both cases. Thus: log2x = ln x / ln 2 or: log2x = log10x / log102
What is log base 3 of (x plus 1) log base 2 of (x-1)?
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
