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    Hey mighty TSR math forum

    I need to retake C2 soon. My logs are sluggish right now, i need to change that! :teeth: here is the question (Btw I have the laws in my head, i just need a push in the right direction):


    Solve 5^(2x) - 12[5^x] + 35 = 0

    I get mixed up with these. I can do the simple logs fine!

    Thanks!
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    If I rewrite 5^(2x) as (5^x)^2, does that help?

    Further hint (only open once you've thought about it):

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    No really, think about it!
    Spoiler:
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    Let u = 5^x
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    5^2x = 5^x x 5^x

    Can you see the quadratic?
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    right. I got it!!

    I did try (5^x)^2 before.

    Now i see it!

    you eventually get (x-5)(x-7)

    I feel less of a crazy scientist now

    Is there a rule with these though? Say there wasnt a 5^(2x) but say a 4^(2x) how would you use logs to solve a ******* like this?
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    (Original post by yurihammo)
    Is there a rule with these though? Say there wasnt a 5^(2x) but say a 4^(2x) how would you use logs to solve a ******* like this?
    In the 4^(2x) case, it's just (4^x)^2...

    And no, you don't get (x-5)(x-7) = 0, you get (u - 5)(u - 7) = 0, which you then need to solve for u and THEN solve for x.
 
 
 
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