Solve the Logarithmic Equation with Radicals | Math Olympiad Training

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Solve the Logarithmic Equation with Radicals | Math Olympiad Training

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[mahalakshmiganapathy]

Very nice, problem and its explanation

[sameerqureshi-khcc]

A seemingly difficult question solved quickly! That's our Premath 😊👍

[MichaelRothwell]

A nice disguised quadratic equation.
The first step, squaring both sides of the equation, could introduce extraneous solutions, namely with log x<0. Therefore, after this step, you need to make the proviso that log x>0 . Luckily both solutions satisfy log x>0, so no extraneous solutions were obtained.

[HappyFamilyOnline]

Amazing step by step explanation👍
Thanks for sharing😊😊

[nirupamasingh]

Vvvv nice explanation. V nice question

[fevengr]

As soon as I squared both sides I recognized the quadratic equation. Nice problem! Thanks!

[farshadfattahi]

Your method is really nice. Thanks a lot, Professor.

[TVS...l-_-l...]

only one word:

COOL !!!!

[math]

this solution was very well explained, love the problem

[parthacreation]

Dear sir please take questions from various differential equations derivatives and integration topics .

[Skank_and_Gutterboy]

Yes! This is what I did, I love me some substitution.

[satrajitghosh]

Writing y for log (x) one gets
y^2 = 13y/6 -1
or 6*y^2 - 9y -4y +6 = 0
or (2y -3)(3y - 2) = 0
or log(x) = 3/2, 2/3
Default base of the logarithm being 10, x = 10^(3/2), 10^(2/3)

[susennath]

I feel my JEE preparation era.
that was 1994

[legendarymathematician]

Sir pls increase frequency of your weakly videos 🙂, as I enjoy them

[j.r.]

I left the fraction in my quadratic equation and then solved it by completing the square, just because I don't get as much practice doing that. :)

[stickmanbattle]

Solve within seconds with the same solution.

[tumak]

...where's the check? When squaring a radical equation new roots might be introduced, so any produced solutions must be checked. Basic math. Cheers

[electromagnezone]

In this channel, the 'log' is common logarithm instead of natural logarithm...

[ΜιχάληςΝτόστας]

Shouldn't you define at first that logx^(13/6) - 1 >= 0

[krislegends]

I eventually got the answer, but this was after I made the mistake of squaring x instead of squaring logx.