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How do you solve log base 2 of x - 3 log base 2 of 5 equals 2 log base 2 of 10?
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
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]
Are there integers a and b that satisfy the equation of log 3 to the base 2 equals a divided by b?
No, log3 to the base 2 is irrational.
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!!!!!
How do you calculate the exponent y if the base is 2 and the answer is 50?
If 2y = 50 then y*log(2) = log(50) so that y = log(50)/log(2) = 5.6439 (approx). NB: The logarithms can be taken to any base >1.
