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myininaya

  • 2 years ago

for supercrazy: how to solve \[3^{4x-7}=4^{2x+3}\]

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  1. myininaya
    • 2 years ago
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    Do you remember \[\ln(x^r)=r \ln(x)\]

  2. myininaya
    • 2 years ago
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    i got stood up

  3. satellite73
    • 2 years ago
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    lol

  4. amistre64
    • 2 years ago
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    those loger-epo rules make more sense when defined by calculus

  5. vengeance921
    • 2 years ago
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    @Myini-Did I do something? o.o Sorry I will delete my comment.

  6. myininaya
    • 2 years ago
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    No this person wanted to know how to do this problem in chat and I said was going to post it here and she never came. :(

  7. vengeance921
    • 2 years ago
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    @Myini-You did your work nevertheless, shows you're a sincere person :)

  8. mkone
    • 2 years ago
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    Just to practice latex \[\begin{align}3^{4x-7}=4^{2x+3} &<=> (4x-7)ln(3) &&=(2x+3)ln(4) \\ &<=> x4ln(3)-7ln(3) &&= x2ln(4)+3ln(4) \\ &<=> x(4ln(3)-2ln(4)) &&= 3ln(4)+7ln(3) \\ & <=> x &&={ 3ln(4)+7ln(3) \over 4ln(3)-2ln(4)} \\ or\ & x={ln(4^3.3^7) \over ln(3^4.4^2) }&& = {ln(139968)\over ln(1296)}\approx 1.65\end{align} \]

  9. myininaya
    • 2 years ago
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    Pretty mkone!

  10. myininaya
    • 2 years ago
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    well i think i see something just a little off on your solution

  11. myininaya
    • 2 years ago
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    \[4 \ln(3)-2 \ln(4)=\ln(3^4)-\ln(4^2)=\ln(\frac{3^4}{4^2})\]

  12. mkone
    • 2 years ago
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    I'm not getting what's off is it the presentation. If so I agree \(ln({3^4 \over 4^2}) \) looks better but then what about \(2ln({3^2 \over 4}) \) ?

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