Evaluate this Log Expression In 1 Minute | NO Calculators Allowed

[ramanivenkata]

Highly impressive. I am not exposed to Logrithm questions of this type. Learnt it from your Tutorial.
Thank you Professor.

[aashsyed]

Very nicely done premath thanks so much

[JSSTyger]

x = log3(2) means 3^x=2
y = log9(4) means 9^y=4 means 3^(2y) = 4 and 3^y = 2
z = log27(8) means 27^z=8 means 3^(3z) = 8 and 3^z = 2
So we have 1/log3(2)+2/log3(2)-3/log3(2) = [1/log3(2)][1+2-3] = 0

[dhrubajyotisarma]

Great question and golden solution sir

[nicogehren]

very helpfull use of log properties sir thanks

[aakarshansubedi]

I never miss your video . It is just like gold for me . Personally. Love from nepal🇳🇵🇳🇵🇳🇵

[sie_khoentjoeng]

Since 1/log(b)a = log b / log a, then 1/log(3)2 = log 3 / log 2.
Similar
2/log(9)4 = 2.log 9/log 4 = 2.log 3^2/log 2^2 = 2.2 log 3/(2.log 2) = 2.log 3 / log 2
And
3/log(27)8 = 3.log 27 / log 8 = 3..log 3^3 / log 2^3 = 3.3.log 3 / (3.log 2) = 3.log 3 / log 2.
If X = log 3 / log 2, the the equation becone:
X + 2X - 3X = 0

[kennethbautista]

We can use ln,
Log b base a= x ; a^x =b ; x= ln b/ln a
1/log 2 base 3= (ln 3 /ln 2)
2/log 4 base 9 = (2ln 3 / ln 2)
3/log 8 base 27= (3ln 3/ln 2)
(ln 3/ln 2)+(2ln 3 /ln 2) - (3ln 3/ ln 2) = (3ln 3/ln 2)- (3ln 3/ln 2) = 0
Answer: 0

[ramaprasadghosh]

1/.(log 2) to the base 3
+ 2/(2.(log 2) to the base 9)
- 3/ (3.(log 2) to the base 27)
=(log 3) to the base 2
+ (log 9) to the base 2
- (log 27) to the base 2
= (1+2-3)*(log 3) to the base 2
= 0

[danilopisani]

could you please make a video showing us what kind of path would you recommend to a person who would like to relearn math?

[highlyeducatedtrucker]

2nd day in a row that you've taught me a mathematical trick I didn't know. And I have a degree in chemical engineering with a minor in mathematics, so I know a lot of mathematical tricks. But I had never heard of how the reciprocal of a logarithm switches the base and the argument. That's a new one on me! Thanks for teaching me something cool today!

[devondevon]

Answer =0
1/log 3 2 + 1/log 3^2 2^2 - 1/log 3^3 2^3
1/log 3 2 + 2/ 2 log 3 2^2 - 3/3 log 3 2^3
1/ log 3 2 + 1/log 3 2^2 - 1/log 3 2^3 (the 2 and 3 in the denominator cancel the 2 and 3 in the numerator)
1/log 3 2^1 + 1/log 3 2^2 -1/1og 3 2^3
1/log 3 2^3- 1/log 3 2^3 =0

[tgx]

log_3 (2)=ln2/ln3, similar log_9 (4)=ln4/ln9=ln2/ln3, log_27 (8)= ln 8/ln27=ln2/ln3

[fsyi]

converting log(9)4 and log(27)8 to log(2)3 and evaluate will be the quickest way

[India-jqpi]

Today my sir also teach about logarithms only. What a coincidence 😮😁

[randaya]

I did use the change of base formula, couple with adding and subtracting fractions to get the numerator to zero, thus the answer is zero. There are some contents in your video that I had never heard of, and it was fun to see it. I like this video.

[highpath]

Every time I did maths I ended up with 0 = 0 . I thought half the time I was wrong.

[petereziagor]

1 minute into the video and I can already see the answer 😁

[Jared]

I forgot the golden logarithmic power rule.

[danilopisani]

to solve this type of exercise what degree of knowledge of mathematics do you need?