Hello :P

How do i simplify this?:

log4(16)*log4(4)

And this:

f(x)=xsin^(-1)x

(Happy) thanks...!

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- Sep 20th 2009, 12:21 PMMissWonderxsin^-1 x
Hello :P

How do i simplify this?:

log4(16)*log4(4)

And this:

f(x)=xsin^(-1)x

(Happy) thanks...! - Sep 20th 2009, 12:27 PMKasper
Welcome to MHF!

2 rules very handy for this type of problem are;

$\displaystyle log_{a}a=1$

$\displaystyle log_{a}(b^c)=c \cdot log_{a}b$

So we know $\displaystyle log_{4}4 = 1$

and we can simplify $\displaystyle log_{4}16=log_{4}(4^2)=2log_{4}4$.

Therefore,

$\displaystyle (log_{4}16)(log_{4}4)$

$\displaystyle =(log_{4}(4^2))(1)$

$\displaystyle =(2 \cdot log_{4}4)(1)$

$\displaystyle =(2)(1)(1)$

$\displaystyle =2$

Does this help? - Sep 20th 2009, 12:40 PMKrizalid
- Sep 20th 2009, 12:42 PMMissWonder
Wee yes ;D I did not know the first part with "loga(A)=1".... Thanks kasper, you're the man!... (Heart)

What about "xsin^(-1)x"? - Sep 20th 2009, 12:44 PMMissWonder
$\displaystyle x(sin^-1)x$... better? I really don't know how to write it in a good way :(

- Sep 20th 2009, 12:54 PMKasper
- Sep 20th 2009, 01:00 PMMissWonder
Yeah :P I really can't figure it out either! Maybe differentiate it is sufficient?..

- Sep 20th 2009, 01:04 PMKasper
It's not so much that it can't be figured out, there is just no way to simplify that expression further.

Keep in mind that differentiation and algebraic simplification are two very different things!

Differentiating will bring out an entirely new function that is not at all equivalent to the original. - Sep 20th 2009, 01:10 PMMissWonder
I know.. But I can't simplify it.. So maybe differentiate it, and then simplify it?

- Sep 20th 2009, 01:14 PMKasper
No, no, because simplification is essentially rewriting a function such that it is equivalent to the original function and so it looks nicer and is easier to work with.

If you differentiate, you lose that equivalency to the original function; so nothing to do with differentiation can be considered an algebraic simplification. Sure you *might* be able to simplify the derivative, but it would not be a simplification of $\displaystyle (x)sin^{-1}(x)$. - Sep 20th 2009, 01:27 PMMissWonder
OMG! I just wrote an e-mail to my mathteacher.. She wrote that a differentiation is sufficient

- Sep 20th 2009, 02:57 PMmr fantastic
Perhaps your math teacher knew the exact instructions that came with the expression f(x)=x sin^(-1)x. Because we certainly didn't, despite several increasingly frustrated attempts to find out what they were.

Next time post all the instructions given in the question.