December math problem

As the description says!

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terminator
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December math problem

Post by terminator »

3^log(2x)=4x


x=?

(Exact answer, not decimal form)
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madmaniacal1
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Re: December math problem

Post by madmaniacal1 »

Term, I genuinely think you have a "math problem" here.
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Re: December math problem

Post by WARBRINGER »

Weget it, Einstein! :lol:
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terminator
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Re: December math problem

Post by terminator »

madmaniacal1 wrote: Thu Dec 26, 2019 10:36 pm Term, I genuinely think you have a "math problem" here.
hehehe :)


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
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LostKnight
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Re: December math problem

Post by LostKnight »

terminator wrote: Fri Dec 27, 2019 8:05 am
madmaniacal1 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 ....
Hmmmm ... me thinks he doth aßume^² much.

:D :D
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