Olympiad Mathematics | Two Methods to Solve System of Exponential Equations | Math Olympiad Training

Can you solve the given system of exponential equations to find the value of abcd? Learn two different approaches to solve this problem.

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Olympiad Mathematics | Two Methods to Solve System of Exponential Equations | Math Olympiad Training

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[mathrapper]

It may look hard at first but with your explanation, it is very easy. Great video.

[davidfromstow]

Hadn't got a clue where to start, but, as always, you make it so understandable

[normanc]

Thank you for your video. I used the 2nd method. I appreciate your 1st method of substitution.

[souldreamer]

First, let me say I admire your videos very much. There is nobody on YouTube that explains things as clearly as you. Now, a question: I don’t understand the Log thing 100%. In Excel and Apple’s Numbers, log is a function that requires TWO variables - the “positive number” and the “base”. If there is only one variable like in this example (which I presume is meant to be the “positive number”), how does one know which base to use?

[HappyFamilyOnline]

Amazing presentation👍
Thanks for sharing🌺

[massimogranzotto]

Ottimo video professore come sempre.
Grazie

[bigm]

The first method was elegant. I wouldn’t say that method two was clumsy……just not elegant. Thanks Professor!🍺🥂

[Copernicusfreud]

I solved it using the first method. Yay!

[devondevon]

The answer can also be expressed as log base (2) 12^1/3 or log base 2 cube root of 12

3^2 = 2^3a since 9=3^2
but 3^2 = 10^1/b (raised both sides to the 1/b)
Hence 2^3a = 10^1/b
Hence 10 = 2^3ab (raised both sides to b)
10^1 = 11^1/c (raised both sides to 1/c)
Hence 11^1/c = 2^3ab
11 = 2^3abc (raised both sides to c)
but since 11^d = 12 ( given), then
11 = 12^1/d (raised both sides to the power of 1/d
Hence 12^1/d = 2^3abc
12 = 2^3abcd
log 12 = 3abcd log 2
log 12/log 2 = 3abcd
1/3 (log 12/log 2)= abcd (divide both sides by 1/3)
log 2 12^1/3 Answer or 1.1947

[kennethgoldberg]

When getting the approximation, make you include the right parenthesis after the 12
Log(12)÷log(8)

[larisamedovaya]

I did in different way: 2^3a= 9; 9^b=10; I substitute 9 in 9^b by 8^a; and so on I got 8^(abc)=12; then log (base 2) of 8^a*b*c*d=3abcd; 3abcd= log(base 2)4+ log(base 2) of3; 3*abcd= 2+log( base 2) 3; abcd=2/3)+ (log(base2)3)/3 ; if you want to use Table of logarithm of base 10 or natural logarithm just use formula change of base : logarithm (base b) of a= logarithm ( base 10) or (base 10) of “a”/
logarithm (base 10) or(base “e”) of “b”

[ilyashick]

Ok, 12 is equal 11+1 take both side log base 11 and get d*log base 11 of 11.result d=1.similar c=1, next 3sqr2b=3sqr2+1, then log on base 3 result 2b=2, b=1, and last 2sqr3a = 2sqr3+1 get log base 2, 3a*log base 2 of 2 = 3* log base 2 of 2 +log base 2 of 1. Result 3a=3 then a=1.

[illyriumus]

I solved it the 2nd way immediately. But, your method one wins for creativity!

[alster]

Wow, 2 chain rules. The 2nd method is way easier though.


On a side note, Log 10 = 1 for method 2

[MrLidless]

Use natural logs (ln) to get a common base throughout. *NEVER* log. SO much quicker.

[ralkadde]

One attempt, solve it from the top: Let us first bring the a, b, c, d on one side: (eq1) a = log9 / log8, (eq2) b = log10 / log9, (eq3) c = log11 / log10, (eq4) d = log12 / log11.
Solution: a*b*c*d = ( log9 / log8) * ( log10 / log9) * (log11 / log10) * (log12 / log11) = log12 / log8 = log₈(12) [This is log 12 base 8, subscripted digit 8 hard to see.]

[seroujghazarian]

2^(3a)=9
3^(2b)=10
(3^2)^b=10
9^b=10
(2^(3a))^b=10
2^(3ab)=10
10^c=11
(2^(3ab))^c=11
2^(3abc)=11
11^d=12
(2^(3abc))^d=12
2^(3abcd)=12
3abcd=2+log2(3)
abcd=(2+log2(3))/3

[ACheateryearsago]

Hey can we say that log12÷log8= log4? Thank you!

[nrellis]

in the second method, once you've shown how a=(log 9)/(log 8), it's okay to write b= (log 10)/(log 9), c= ... etc without going through all the steps, we can take it as read

[abeonthehill]

Can anyone help with this one, ……….?

9^y + 18^y =36^y ……….this one i could not solve from my book !