3^log(2x)=4x
x=?
(Exact answer, not decimal form)
December math problem
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- terminator
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December math problem
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- madmaniacal1
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Re: December math problem
Weget it, Einstein!
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- terminator
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Re: December math problem
hehehemadmaniacal1 wrote: ↑Thu Dec 26, 2019 10:36 pm Term, I genuinely think you have a "math problem" here.
Ill give you a hint
Since many know the general rule in which log(a^b)=a*log(b) we can see a VERY interesting thing occur if we did this.....
log(a)*log(b)....You can see that in this case it means......log(a)*log(b)=log(a^log(b))=log(b^log(a))....
Which is quite interesting as it leads to this .... log(a^logb)=log(b^loga)....remove log from both sides.....a^logb=b^loga
This will be very helpful in solving this
Terminator-2000+() archer
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My quest is to promote the future and prevent ill history from repeating itself.
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My quest is to promote the future and prevent ill history from repeating itself.
- LostKnight
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Re: December math problem
Hmmmm ... me thinks he doth aßume^² much.terminator wrote: ↑Fri Dec 27, 2019 8:05 amIll give you a hintmadmaniacal1 wrote: ↑Thu Dec 26, 2019 10:36 pm Term, I genuinely think you have a "math problem" here.
Since many know the general rule in which ....
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