How can the change-base identity be used to prove this equation?

[tahayassen]

Homework Statement

http://img843.imageshack.us/img843/3826/help3c.png

Homework Equations

Not applicable.

The Attempt at a Solution

http://img810.imageshack.us/img810/5577/help2n.png

Can anyone prove this by evaluating the left side and right side independent of each other?

http://img827.imageshack.us/img827/9757/helpdi.png

Can someone explain why this is incorrect? I'm so confused.

 

[Ray Vickson]

Homework Statement

http://img843.imageshack.us/img843/3826/help3c.png

Homework Equations

Not applicable.

The Attempt at a Solution

http://img810.imageshack.us/img810/5577/help2n.png

Can anyone prove this by evaluating the left side and right side independent of each other?

http://img827.imageshack.us/img827/9757/helpdi.png

Can someone explain why this is incorrect? I'm so confused.

$$2^{3^4} = 2^{81} = 2417851639229258349412352, \mbox{ while } 2 \times 3 \times 4 = 24.$$ In fact, $$((x^a)^b)^c = x^{a b c},$$ as you have recognized, but that is _very_ different from $$x^{a^{b^c}},$$ which you are confused about. Look at the brackets: in $((x^a)^b)^c$ we first compute $x^a,$ then take the bth power of that, then take the cth power of the result. In x^[a^(b^c)] we first compute b^c, then raise a to that power, then raise x to the result.

RGV

 

[tahayassen]

$$2^{3^4} = 2^{81} = 2417851639229258349412352, \mbox{ while } 2 \times 3 \times 4 = 24.$$ In fact, $$((x^a)^b)^c = x^{a b c},$$ as you have recognized, but that is _very_ different from $$x^{a^{b^c}},$$ which you are confused about. Look at the brackets: in $((x^a)^b)^c$ we first compute $x^a,$ then take the bth power of that, then take the cth power of the result. In x^[a^(b^c)] we first compute b^c, then raise a to that power, then raise x to the result.

RGV

Interesting. Thanks for the clarification.

If anyone is wondering what I meant by evaluating each side independent of each other, I meant evaluating the left-hand side, and then evaluating the right-hand side, and then after they look exactly the same, you can conclude that the left-hand side equals the right-hand side. Unfortunately, this is the only way my school accepts proofs.

 

[Simon Bridge]

Operating only on the LHS: you first prove the relation to change base and then use that identity, vis:$$LHS=\log_{(a^b)}(c^d) = \frac{\log_a(c^d)}{\log_a(a^b)} = \frac{d\;\log_a(c)}{b}=RHS$$...

(If you have already done the proof [change-base identity], i.e. in class, you can just refer to it; and if you haven't, you can look it up.)

Aside: it is probably worth your while learning LaTeX rather than posting images.
It helps us too since we can then copy and paste from your post to edit your equations.

 

[tahayassen]

Operating only on the LHS: you first prove the relation to change base and then use that identity, vis:$$LHS=\log_{(a^b)}(c^d) = \frac{\log_a(c^d)}{\log_a(a^b)} = \frac{d\;\log_a(c)}{b}=RHS$$...

(If you have already done the proof [change-base identity], i.e. in class, you can just refer to it; and if you haven't, you can look it up.)

Aside: it is probably worth your while learning LaTeX rather than posting images.
It helps us too since we can then copy and paste from your post to edit your equations.

Thanks! That helped a ton! :)

@LaTeX: I looked at the tutorial, and it looked rather complicated... I've been using Daum Equation Editor (a chrome extension) and it has a beautiful interface. I'll learn LaTeX as soon as this semester is over though.

http://img860.imageshack.us/img860/1480/32446906.png