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Suppose aneq 0. Compute log 2a 2b in terms of a and b.?
Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]
Solve ln y equals xln e?
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
What is log of 1 to base 10?
It is zero
What is log base 5of 125?
log(5)125 = log(5) 5^(3) = 3log(5) 5 = 3 (1) = 3 Remember for any log base if the coefficient is the same as the base then the answer is '1' Hence log(10)10 = 1 log(a) a = 1 et.seq., You can convert the log base '5' , to log base '10' for ease of the calculator. Log(5)125 = log(10)125/log(10)5 Hence log(5)125 = log(10) 5^(3) / log(10)5 => log(5)125 = 3log(10)5 / log(10)5 Cancel down by 'log(10)5'. Hence log(5)125 = 3 NB one of the factors of 'log' is log(a) a^(n) The index number of 'n' can be moved to be a coefficient of the 'log'. Hence log(a) a^(n) = n*log(a)a Hope that helps!!!!!
What is the logarithm of log -1?
Assuming you are asking about the natural logarithms (base e):log (-1) = i x pithereforelog (log -1) = log (i x pi) = log i + log pi = (pi/2)i + log pi which is approximately 1.14472989 + 1.57079633 i
