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myininaya
 4 years ago
for supercrazy:
how to solve \[3^{4x7}=4^{2x+3}\]
myininaya
 4 years ago
for supercrazy: how to solve \[3^{4x7}=4^{2x+3}\]

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1Do you remember \[\ln(x^r)=r \ln(x)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0those logerepo rules make more sense when defined by calculus

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@MyiniDid I do something? o.o Sorry I will delete my comment.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1No 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. :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@MyiniYou did your work nevertheless, shows you're a sincere person :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just to practice latex \[\begin{align}3^{4x7}=4^{2x+3} &<=> (4x7)ln(3) &&=(2x+3)ln(4) \\ &<=> x4ln(3)7ln(3) &&= x2ln(4)+3ln(4) \\ &<=> x(4ln(3)2ln(4)) &&= 3ln(4)+7ln(3) \\ & <=> x &&={ 3ln(4)+7ln(3) \over 4ln(3)2ln(4)} \\ or\ & x={ln(4^3.3^7) \over ln(3^4.4^2) }&& = {ln(139968)\over ln(1296)}\approx 1.65\end{align} \]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1well i think i see something just a little off on your solution

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[4 \ln(3)2 \ln(4)=\ln(3^4)\ln(4^2)=\ln(\frac{3^4}{4^2})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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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