Solve and Check Logarithmic Equation with Different Bases | Math Olympiad Training

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Solve and Check Logarithmic Equation with Different Bases | Math Olympiad Training

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[SuperYoonHo]

1st!!! Thanks so much for the awesome vid! I got in right in my head it is log2 times log 3 all overy log6. Thanks for your time and it is a hounor to be first haha. Take good care and see you my friend.

[mohamedinsaf]

2^(log3/log6)
3^(log2/log6)
Are also the answers

[dreael]

The same is also possible using the natural log, i.e. ln(a)/ln(2)+ln(a)/ln(3)=1 at the beginning and ending with a=e^(ln(2)*ln(3)/ln(6)) as the end result.

[owlsmath]

Good one! I got 2^log_6 (3) which is the same. I was struggling trying to find a nicer solution!!!

[JSSTyger]

OK so 2^x = a and 3^y = a. We have also that x+y = 1. 2^(1-y) = 3^y. From this, we get 2 = 6^y. So if 3^y = a, then a = 3^(log_6(2)), which comes out to about 1.53.

[HappyFamilyOnline]

Beautiful solution👍
Thanks for sharing💕💕

[andymci]

Another great problem solved by the master! always start these problems with the mindset that calculators are "not allowed", or that it is somehow cheating. Consequently, i spend far more time trying to get a solution than i need to. Maybe a little hint at the start of a video that calculators are ok would be helpful to us. Otherwise, love this channel....i dont know how you think them up.!

[WolfgangKais]

Thisa coulda be simplifieda to
a = 2^log_6(3) = 3^log_6(2).

[MichaelRothwell]

Nice problem and solution.

It turns out that the simplest forms of the answer are obtained by using change of base of the logs to base 6 (this is because the log6 then becomes log₆6=1):

log₂a+log₃a=1
log₆a/log₆2+log₆a/log₆3=1
log₆a(1/log₆2+1/log₆3)=1
log₆a((log₆2+log₆3)/(log₆2 log₆3))=1
log₆a×log₆6=log₆2 log₆3
log₆a×1=log₆2 log₆3
log₆a=log₆2 log₆3
So
a=6^(log₆2 log₆3)
Alternate forms of the answer:
a=(6^log₆2)^log₆3=2^log₆3
a=(6^log₆3)^log₆2=3^log₆2