keketsu 3 years ago Simplify: /[2x(2x)^(x-1) + (ln (2x))(2x)^x/] I'm particularly interested in how to work with /[(2x)^(x-1)/]. thanks.

1. vengeance921

$2x(2x)^{x-1}+(Ln(2x))(2x)^{x}$ Is this the question?

2. amistre64

$a^{(b+c)}=a^b * a^c$

3. vengeance921

They have a natural logarithm there, right?

4. amistre64

yes, but it looks to just care about factoring out the 2x

5. keketsu

Yes, vengeance, sorry about the extra brackets. And yes, amistre64, you gave me the rule I was looking for. Thanks guys. How do I mark these equations so that the system recognizes them as math problems, btw?

6. amistre64

$(2x)^1*(2x)^{(x-1)}=(2x)^{(x-1+1)}$

7. amistre64

latex i assume

8. vengeance921

@Keketsu=Can you just post the real question please? Lol

9. SchoolSlacker

2x(2x)^{x-1} sorry for asking my own question in your question but i just wanted to know with this would you do the factorisation with the outer number first or the indices with the brackets?

10. amistre64

a demonstration to type in latex would be futile since the system would parse it as latex and not a demonstration :)

11. keketsu

@Vengeance921, what you posted is the equation, but it is dealing with the (2x)^(x-1) that I was having trouble with. So that's (2x^x) * (2x^-1) or (2x^x)/2x?

12. amistre64

School, there is no real rule for procedure; only that it makes mathical sense in the end :)

13. amistre64

$\frac{2x*2x^x}{2x}$is one way to present it, yes

14. SchoolSlacker

Thanks i just saw it in the equation, and i had no idea :P

15. keketsu

Wonderful. Thanks for the help.

16. amistre64

youre welcome