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Can you multiply two irrational logarithms to get a rational answer?
Yes. Take any rational number p. Let a = any number that is not a power of 10, so that log(a) is irrational. and let b = p/log(a). log(a) is irrational so 1/log(a) must be irrational. That is, both log(a) and log(b) are irrational. But log(a)*log(b) = log(a)*[p/log(a)] = p which is rational. In the above case all logs are to base 10, but any other base can be used.
What is the difference between an anti-log and an exponential?
let x and y be two numbers ex = y log y = x antilog x = y
How do you half a number by using only powers?
You can't: let suppose y the power of x to obtain such a result then xy=x/2 then xy-1=1/2 (y-1) log (x) = - log(2) (if x is a positive number) y-1 = -log(2)/log(x) y = 1 - log(2)/log(x) So log function must also being used!
What is the log prosedure?
I assume you are asking how to solve a logarithmic equation. Well let's quickly review what the log function is: for the equation log(x)=y, we are saying that 10^y=x. So once you have isolated the logarithm, take the value of the base, raise it to the nth power (when 'n' is the value that the function is equal to) and set that equal to the value inside of the log.
Log 2 plus log 4 equals log 2x?
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
