Thread: Exponential problems

I think I did these correctly but before I do anymore of the same problems I was wondering if you guys could check them for me


Simplify with no radicals or negative exponents:


1. (x(squareroot 16xt^2))^2

I'm not certain what sign to use on the keyboard for squareroot symbo

Well this is what I did.

(x^2)(16xt^2)

16x^3t^2


2. (3ab)^-1(a^2b^-1)^2

(a^2)(b^-2) / (3ab)

a^2 / 3ab^3

a / 3b^3


Any help or advice would be great

Originally Posted by JonathanEyoon

I think I did these correctly but before I do anymore of the same problems I was wondering if you guys could check them for me


Simplify with no radicals or negative exponents:


1. (x(squareroot 16xt^2))^2

I'm not certain what sign to use on the keyboard for squareroot symbo

Well this is what I did.

(x^2)(16xt^2)

16x^3t^2

this looks fine. if you can't use LaTex, use sqrt() for the square root symbol. so you would type [x*sqrt(16x(t^2))]^2

2. (3ab)^-1(a^2b^-1)^2

(a^2)(b^-2) / (3ab) ...

the power you have for a in the numerator is not correct

Originally Posted by Jhevon

this looks fine. if you can't use LaTex, use sqrt() for the square root symbol. so you would type [x*sqrt(16x(t^2))]^2

the power you have for a in the numerator is not correct

Ok the first one is fine =)

Let me see if I can redo the second one.

(3ab)^-1((a^2)(b^-1))^2

From here I will

(a^4)(b^-2) / (3ab)

(a^4) / (3ab^3)

a^3 / 3b^3


Does that look better?

Originally Posted by JonathanEyoon

Ok the first one is fine =)

Let me see if I can redo the second one.

(3ab)^-1((a^2)(b^-1))^2

From here I will

(a^4)(b^-2) / (3ab)

(a^4) / (3ab^3)

a^3 / 3b^3


Does that look better?

looks good. good job

Originally Posted by Jhevon

looks good. good job

Thanks!

Another question


This problem

(e^kt)(e^3)(e) = e^4kt


Am I right?

Originally Posted by JonathanEyoon

This problem

(e^kt)(e^3)(e) = e^4kt

So it is

This requires the power property

Originally Posted by Krizalid

So it is

This requires the power property

Yea that's the problem. Doh I had it wrong then. So instead it should be

e^3+kt?

Originally Posted by JonathanEyoon

Yea that's the problem. Doh I had it wrong then. So instead it should be

e^3+kt?

nope.

remember,

Originally Posted by Jhevon

nope.

remember,

oopsie forgot about that . Ok my 3rd attempt!!


So the final answer should be e^4+kt?

Originally Posted by JonathanEyoon

oopsie forgot about that . Ok my 3rd attempt!!


So the final answer should be e^4+kt?

yes. but type e^{4 + kt} so that we know the kt is not separate from the e

Originally Posted by Jhevon

yes. but type e^{4 + kt} so that we know the kt is not separate from the e

ok!! I always forget to correctly use the parentheses . I'll try to take more notice of them. Thanks alot guys!!