First we have: 8 Ln(e^a) + 9 ln(B^e) 8a Ln(e) + 9 ln(B^e) 8a + 9ln(B^e) Anybody know what to do from here?
Follow Math Help Forum on Facebook and Google+
Originally Posted by Gfrant7 First we have: 8 ln(eA) + 9 ln(Be) 8a Ln(e) + 9 ln(B^e) 8a + 9ln(B^e) Anybody know what to do from here? wait, i'm confused. it seems you are working with powers in the second line, yet in the first line you only have products. which is it? do you mean ? if so, you should indicate powers by using "^" just as you did in the second line
Hello, The only thing you can do is ln(b^e)=e*ln(b). What was it for ?
Sorry Jhevon, I didn't have it written correctly in my initial post. That is what I meant though.
Originally Posted by Gfrant7 Sorry Jhevon, I didn't have it written correctly in my initial post. That is what I meant though. well, in that case, follow Moo's post. you can't really to anything more than changing to if you were told the values of ln(B) at some point, you can plug it in, if not, just leave it
In that case we would get 8ln(e^a) + 9ln(b^e) which would simplify to 8a + 9eln(b) which webassign is telling me is the incorrect answer. damn this problem lol
Oh ! I think I can see what you're asked. use these rules and and --> --> ----->
Originally Posted by Moo Oh ! I think I can see what you're asked. use these rules and and --> --> -----> Oh! haha, ok! that's makes sense you should have told us you were trying to simplify into a single logarithm, Gfrant7
Thanks! would it be ln(e^8a) + ln(b^9e) = ln((e^8a)*(b^9e))
Originally Posted by Gfrant7 Thanks! would it be ln(e^8a) + ln(b^9e) = ln((e^8a)*(b^9e)) yes, but make sure your syntax is correct if you are typing it. you might need the power in brackets, otherwise it might interpret b^9e as when you mean . and so on...
View Tag Cloud