How do I condense the expression 3/2ln5t^6-3/4lnt^4?

[ohchelsea]

i have to condense this expression:

3/2ln5t^6-3/4lnt^4

i may have done this in the wrong order but so far I've gotten to this point:
(3/2ln5t^6)/(3/4lnt^4)am i doing this right? is that all i have to do?

 

[cristo]

Remember the rule: logx^a=alogx

 

[ohchelsea]

so what do i do about those fractions and the exponents that are already there?

 

[cristo]

Well, start from the top again, [since you use the incorrect division of logs rule; the correct one is that loga-logb=log(a/b)].

Firstly, tidy up the expression and remove the exponents, so, what would $$\frac{3}{2}\ln 5t^6$$ simplify to, using the rule I gave in post 2? Similarly, what would the second term become?

Then use the division property I give above to combine the two into one logarithm.

 

[HallsofIvy]

i have to condense this expression:

3/2ln5t^6-3/4lnt^4

i may have done this in the wrong order but so far I've gotten to this point:
(3/2ln5t^6)/(3/4lnt^4)


am i doing this right? is that all i have to do?

What exactly do you mean by "condensing" the expression? Writing it as a single expression. Your basic idea is right but you are misremembering a "law of logarithms".
ln a- ln b= ln (a/b), NOT "ln(a)/ln(b)". Also, as cristo pointed out, you need to handle those coefficients, (3/2) and (3/4).
$$\frac{3}{2}ln(5t^6)= ln((5t^6)^{3/2}$$

 

[ohchelsea]

okay so i got this:

(ln5t^9)/(lnt^3)

i still don't feel like i did that right =/

 

[cristo]

Sorry, I read that as ln(5t)^6 and then used horrible notation in my post!

okay so i got this:

(ln5t^9)/(lnt^3)

i still don't feel like i did that right =/

You're still using the law of subtraction of logs incorrectly! Also, looking at Halls' post, note that your first term should be $$\ln(5^{3/2}t^9)$$. Then look again at the rule loga-logb=log(a/b)

 

[ohchelsea]

okay i see how i was doing it wrong now...sorry had a dumb moment.

so then would i get
ln(5^3/2*t^9)/t^3) ?

 

[cristo]

Ok, can this be simplified?

[Hint: $$\frac{x^a}{x^b}=x^{a-b}$$ ]

 

[ohchelsea]

ln(5^3/2)t^6 ?

 

[cristo]

Yea, it's all in the argument of the log function though: $$\ln(5^{3/2}t^6)$$

 

[ohchelsea]

yay! thank you so much!