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    How come the 2 just immediately goes away?

    Do you "take logs" to both sides, and therefore cancel out "log2" leaving the algebraic equation behind?
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    (Original post by im so academic)


    Do you "take logs" to both sides, and therefore cancel out "log2" leaving the algebraic equation behind?
    Yes, the 'log2's cancel.
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    (Original post by ChopinLiszt)
    Yes, the 'log2's cancel.
    So if there's an equation with the same base, you can just cancel the base via the logs? I.e. it would work with any similar question?
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    (Original post by im so academic)


    How come the 2 just immediately goes away?

    Do you "take logs" to both sides, and therefore cancel out "log2" leaving the algebraic equation behind?
    You can take logs of both sides and then cancel out log(2), but i don't do it that way.

    You've got an equation where 2^a = 2^b
    For this to be true, then a=b. So you just equate the powers.
    In this case you get 3y = 2(2x + 3), which gives you 3y = 4x + 6.

    For any equation in the form x^a = x^b,  a = b
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    (Original post by F1Addict)
    You can take logs of both sides and then cancel out log(2), but i don't do it that way.

    You've got an equation where 2^a = 2^b
    For this to be true, then a=b. So you just equate the powers.
    In this case you get 3y = 2(2x + 3), which gives you 3y = 4x + 6.

    For any equation in the form x^a = x^b,  a = b
    Thanks for the rule. I never actually knew that.
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    (Original post by im so academic)
    Thanks for the rule. I never actually knew that.
    If you want to understand where this rule comes from, a good way to think about it is to try and solve the following for 'a' and 'b':
    x^a=x^b, where x \not= 1 (otherwise there wouldn't be a unique solution for 'a' and 'b' as 1^a=1^b for all values of a and b over the reals)
    \implies \frac{x^a}{x^b} = 1
    \implies x^{a-b}=1
    Note that x^k=1 if and only if k=0.
    In this case, a-b=0 for this to be true, therefore a=b.

    So essentially, what I'm saying is that if you have two equal bases on both sides raised to particular powers, then those two powers must be equal.
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    (Original post by Farhan.Hanif93)
    If you want to understand where this rule comes from, a good way to think about it is to try and solve the following for 'a' and 'b':
    x^a=x^b, where x \not= 1 (otherwise there wouldn't be a unique solution for 'a' and 'b' as 1^a=1^b for all values of a and b over the reals)
    \implies \frac{x^a}{x^b} = 1
    \implies x^{a-b}=1
    Note that x^k=1 if and only if k=0.
    In this case, a-b=0 for this to be true, therefore a=b.

    So essentially, what I'm saying is that if you have two equal bases on both sides raised to particular powers, then those two powers must be equal.
    Thats a proper derivation there. I derived it using logic. I just thought that the only way x^a can equal x^b, is if a=b. Putting some numbers in that, its evident that 2^4 does not equal 2^3 or 2^5, so it was clear that the only way the equation would work is if the power were equal. :P
    The derivation is a better way of doing it though.
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    (Original post by Farhan.Hanif93)
    If you want to understand where this rule comes from, a good way to think about it is to try and solve the following for 'a' and 'b':
    x^a=x^b, where x \not= 1 (otherwise there wouldn't be a unique solution for 'a' and 'b' as 1^a=1^b for all values of a and b over the reals)
    \implies \frac{x^a}{x^b} = 1
    \implies x^{a-b}=1
    Note that x^k=1 if and only if k=0.
    In this case, a-b=0 for this to be true, therefore a=b.

    So essentially, what I'm saying is that if you have two equal bases on both sides raised to particular powers, then those two powers must be equal.
    Oh that is really helpful. Thank you very much!
 
 
 
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