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

  • Thread starter tahayassen
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In summary: 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.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
  • #1
tahayassen
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1

Homework Statement



http://img843.imageshack.us/img843/3826/help3c.png [Broken]

Homework Equations



Not applicable.

The Attempt at a Solution



http://img810.imageshack.us/img810/5577/help2n.png [Broken]

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

http://img827.imageshack.us/img827/9757/helpdi.png [Broken]

Can someone explain why this is incorrect? I'm so confused.
 
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  • #2
tahayassen said:

Homework Statement



http://img843.imageshack.us/img843/3826/help3c.png [Broken]

Homework Equations



Not applicable.

The Attempt at a Solution



http://img810.imageshack.us/img810/5577/help2n.png [Broken]

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

http://img827.imageshack.us/img827/9757/helpdi.png [Broken]

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

[tex] 2^{3^4} = 2^{81} = 2417851639229258349412352, \mbox{ while } 2 \times 3 \times 4 = 24. [/tex] In fact, [tex] ((x^a)^b)^c = x^{a b c},[/tex] as you have recognized, but that is _very_ different from [tex] x^{a^{b^c}},[/tex] which you are confused about. Look at the brackets: in [itex] ((x^a)^b)^c [/itex] we first compute [itex] x^a,[/itex] 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
 
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  • #3
Ray Vickson said:
[tex] 2^{3^4} = 2^{81} = 2417851639229258349412352, \mbox{ while } 2 \times 3 \times 4 = 24. [/tex] In fact, [tex] ((x^a)^b)^c = x^{a b c},[/tex] as you have recognized, but that is _very_ different from [tex] x^{a^{b^c}},[/tex] which you are confused about. Look at the brackets: in [itex] ((x^a)^b)^c [/itex] we first compute [itex] x^a,[/itex] 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.
 
  • #4
Operating only on the LHS: you first prove the relation to change base and then use that identity, vis:[tex]LHS=\log_{(a^b)}(c^d) = \frac{\log_a(c^d)}{\log_a(a^b)} = \frac{d\;\log_a(c)}{b}=RHS[/tex]...

(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.
 
  • #5
Simon Bridge said:
Operating only on the LHS: you first prove the relation to change base and then use that identity, vis:[tex]LHS=\log_{(a^b)}(c^d) = \frac{\log_a(c^d)}{\log_a(a^b)} = \frac{d\;\log_a(c)}{b}=RHS[/tex]...

(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 [Broken]
 
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1. What does it mean to "prove an identity"?

Proving an identity means showing that two mathematical expressions are equal for all values of the variables involved, using algebraic manipulation and logical reasoning.

2. Why is proving an identity important in science?

Proving an identity is important in science because it allows us to verify the validity of mathematical equations and theories. It helps support and strengthen scientific arguments and conclusions.

3. How do you prove an identity?

To prove an identity, you need to manipulate one side of the equation using algebraic rules and properties until it is equivalent to the other side. This involves simplifying expressions, combining like terms, and using identities and equations that have been proven previously.

4. What are some common strategies for proving an identity?

Some common strategies for proving an identity include starting with the more complex side of the equation, breaking down expressions into simpler forms, using known identities and equations, and paying attention to patterns and symmetries in the expressions. It is also important to make sure that the steps taken are reversible and do not change the overall value of the equation.

5. Can an identity be proven false?

No, an identity cannot be proven false. If an identity is false, it is not considered an identity at all. An identity is a true statement that holds for all values of the variables involved.

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