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.
let x and y be two numbers ex = y log y = x antilog x = y
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!
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) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
Let your teacher do it for u first your teacher has to go to her Edmodo go to students the press remove from group then that's how u delete it.
If you log into your account, go to your profile. Replace the ID number with "1" - Judging from the profile ID of 1, Mr. Borg invented Edmodo and was the first person on the website. -
because they will kicked you out
edmodo
why wont theslap.com let me log out?
What do you do if you forgotten your password to edmodo
You don't get a free group code for edmodo, the free group code for edmodo gets you...
if it wont let u log on wait a day or 2 and it will let u log on
educational
yes
The group code for Edmodo is distributed by teachers so you can join their class.
You first have to have a teacher which gives assignments, via edmodo, then you have to turn it in via edmodo, and get a 90 or higher, thats how you get one of them.OR u can Join the group kbw25j free badges to anyone that joins