Simplify: /[2x(2x)^(x-1) + (ln (2x))(2x)^x/]

I'm particularly interested in how to work with /[(2x)^(x-1)/]. thanks.

Question

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

They have a natural logarithm there, right?

amistre64 · Answer

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

anonymous · Answer

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?

amistre64 · Answer

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

amistre64 · Answer

latex i assume

anonymous · Answer

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

anonymous · Answer

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?

amistre64 · Answer

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

anonymous · Answer

@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?

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

Wonderful.  Thanks for the help.

amistre64 · Answer

youre welcome