Evaluating a Logarithmic Expression in Three Ways

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#ChallengingMathProblems #LogarithmicExpressions

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[SyberMath]

Here's a similar problem per Tanish Sharma's request:
log_20 (10)=x
Find log_5 (2) in terms of x.

[Jha-s-kitchen]

From when I have started watching you, this is the first ever question I solved at my first attempt (and got the (correct) solution too).

Then I watched the video and found my method to be the mix of 1st and 3rd methods.

Thanks, keep making these questions and we will solve everyone of them.

[佐藤広-cp]

From the perspective of the solution and the shape of x, it feels normal to convert the base to a logarithm with a two.
That is, x = log_12 (24) = log_2 (24) / log_2 (12) = (3log_2 (2) + log_2 (3) )/ (2log_2 (2) + log_2 (3)) = (3 + log_2 (3) / (2 + log_2 (3)),
then the rest will only be transformed into log_2 (3)=.

[cube]

That was ez.. Thanks for bringing Mr log back.

[johnlashlee]

Write 12^x = 24 and 2^y = 3. Rewrite as 12^(x-1) = (2*2*3)^(x-1) = 2. Raise both sides to y. Replace all factors of 2^y with 3. Then equate powers to get (2+y)(x-1) = 1. Thanks for bringing Mr. Log back.

[jaimeduncan]

Great video, in the second method is better to use the change rule to base two from the beginning. I did not saw the 3th method pretty original.

[davidseed]

Since the answer uses log_2 use that throughout.
x=log_12(24) =
log_2(24)/log_2(12) =
but
log_2(3)=A (Answer)
So {3+A}/{2+A} =x
Rearrange to give
A={2x-3}/{1-x}

[navghtivs]

Let's call what we look for y = log_2(3),
I call ln(2)=a, ln(3)=b, so y=b/a, or: b = ay;
And: x = (3a+b)/(2a+b);
putting these two together we get a linear equation between x and y without a or b.

[RisetotheEquation]

Method #2 using Change of Base was the way I did it. Method #3 really is just a more sophisticated spin-off of the change of base with a substitution.

[ssa]

Решение длинное и не используются свойства логарифмов.
log24=log(12*2)=log12+log2=
так как основание 12, то
=1+log2=
переход к основанию 2
= 1+ log2/log12=1+1/log(4*3)=
=1+1/(log4+log3)=
=1+1/(2log2+log3)=
=1+1/(2+log3).
Итак, 1+1/(2+log3)=x.
Отсюда 1/(2+log3)=x-1;
2+log3=1/(x-1);
log3=1/(x-1)-2;
log3=(1-2x+2)/(x-1);
log3=(3-2x)/(x-1)
или log3=(2x-3)/(1-x).

[ngavu]

Hello syber i have a question for u here. solve it in next video.for a+b+c=a/(b+c) + b/(c+a) + c/(a+b).calcule P=a^2+1/(b+c) + b^2+1/(a+c) + c^2+1/(a+b)

[mosaaafer]

Thank you for your helpful videos. Can I ask what program u use as digital blackboard?

[partisano]

Thanks you for your videos, diary learning...

[ABHIGAMING-yomy]

Log24/log12=
(log3+3log2)/(log3+2log2)=x
Then subtract 1 from both side
Log2/log3+2log2=x--1
Log3/log2 +2=1/x-1
Log3/log2=3-2x/x-1
This is short way

[oahuhawaii]

log12(24) = x
log12(2) + 1 = x
1/log2(12) = x - 1
2 + log2(3) = 1/(x - 1)
log2(3) = 1/(x - 1) - 2 = (3 - 2*x)/(x - 1)

The above is a better form for the solution than (2*x - 3)/(1 - x) because the numerator and denominator are both positive instead of negative.

[mathsandsciencechannel]

GREAT VIDEO. 100K SUBSCRIBERS VERY SOON

[川の神田の神]

log解法はビジュアルで分かりやすくなります。算数、数学は絵(ビジュアル化)で算数を楽しくできます、とけます、無理に=(イコール)で結ぶのが算数かとおもってます。03, 02, 2022

[bansal]

I challenge you I'm from India I'm giving a question try to do it
In triangle abc ad is the median through a and e is the midpoint of ad be produced meets ac in f prove af=1/3ac without mid point theorem
If you could do it then tell plij

[alanwijaya]

Would you like to tell me the device and software to make this video?

[jimmykitty]

*A little bit tough but interesting one* 💜