1. ## 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

2. 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

3. 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?

4. 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

5. Originally Posted by Jhevon
looks good. good job

Thanks!

Another question

This problem

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

Am I right?

6. Originally Posted by JonathanEyoon
This problem

(e^kt)(e^3)(e) = e^4kt
So it is $e^{kt}\cdot e^3\cdot e$

This requires the power property $a^k\cdot a^m\cdot a^n=a^{k+m+n}$

7. Originally Posted by Krizalid
So it is $e^{kt}\cdot e^3\cdot e$

This requires the power property $a^k\cdot a^m\cdot a^n=a^{k+m+n}$

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

e^3+kt?

8. Originally Posted by JonathanEyoon
Yea that's the problem. Doh I had it wrong then. So instead it should be

e^3+kt?
nope. $e^{kt} \cdot e^3 \cdot e = e^{kt + 3 + 1}$

remember, $e = e^1$

9. Originally Posted by Jhevon
nope. $e^{kt} \cdot e^3 \cdot e = e^{kt + 3 + 1}$

remember, $e = e^1$

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

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

10. 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

11. 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!!