Solve the Logarithmic Equation with Different Bases | Math Olympiad Training

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Solve the Logarithm Equation with Different Bases | Math Olympiad Training

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[mabdinur]

So much easier to use a base change (like change log_7(X) to log_8(X)/log_8(7)) and playing with the fractions of a system of logs with equivalent bases. You get X = 8^(log_8(7)/(1+log_8(7)))

[justabunga]

We can also use the change of base formula as Factor out log(x), which is log(x)(1/log(7)+1/log(8))=1. Divide both sides by 1/log(7)+1/log(8) giving us Solving for x will be or 7^(log_56(8)).

[hansschotterradler]

We definitely need more problems involving logarithms. I must have had a hangover or something else when this stuff was covered in high school.

[devondevon]

x= 2.7325
log 7 x + log 8 x= 1 is saying
7 raised a number = x
and 8 raised to a number also = x but when both numbers are added the total value of both=1
let p = the number 7 is raised to, to =x, hence 1-p is = the number 8 is raised to, to =x as (p)+ (1-p)=1
it is akin to saying if m+ r= 1 pound, hence if m= p pound then r=1-p pound.
hence
7^1-p = x and
8^p= x
hence
7^1-p = 8^p since both = x
1-p log 7 = p log 8 log 7=0.8451 and log 8= 0.9031
0.8451- 0.8451p = 0.9031 p
0.8451 = 1.748p
0.4834134=p
hence p-1 = 0.5165857
hence 7^0.5165857= 8^0.4834134 = 2.7325.. answer

[Mothuzad]

Doing some mental arithmetic, I got 7**(1/(1+log_8(7))), which I then ran through Wolfram to quickly check for equivalence to your solution.

I first used change of base in order to get a common factor of log_8(x), which I factored out leaving (1/log_8(7) + 1), which can also be expressed as (1+log_8(7))/log_8(7). Dividing that factor out of both sides and then raising 8 to that power was enough to isolate x.

[Mekenyejustus]

Great math🙏🙏
Alternatively
(Logx/log7)+(logx/log8)=1
logx(1/log7+1/log8)=1

logx=(log7log8)/(log7+log8)
x=10^{(log7log8)/(log7+log8)}
~2.733

[anatoliy]

It's like attending a math lesson on logarithms with different bases at my home school:)) Thank you so much, sir! God bless you!

[ishaanlohani]

Loved this, I am Preparing for the Indian JOINT ENTRANCE EXAM, and this was really helpful🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[Clyntax]

I directly derived 7^m = x, 8^n = x and m+n=1. Solving for n gives n=mlog_7(8) so that m=1/(1+log_7(8)) = log_56(7). Hence, x = 8^m = 8^log_56(7).

[xyz]

Can also be done by simply express log_8 x as log_7 x / log_7 8, the rest is straight forward without the need for any substitution.

[ruilongsheng]

(log 8+ log 7)*logx =log 7*log8, log x= log7*log8/log 56, x=10^(log 7*log8)/log 56, caculated as: 10^0.4365≈2.7321 very close to the answer 7^0.5166≈2.733

[whaddoiknow]

This is a very nice video, but it raises a broader issue. A lot of people who say they are not "good at maths" would find the approach arbitrary and based on special "tricks." To these people I think it is important to say that solving such problems is a matter of high-level pattern recognition. I do not know how to teach anyone how to recognize patterns except to have seen lots of patterns.

[ronaldnoll]

At least I can write comments here. I was hooked on the critical YouTube posts. That's called censorship. I don't have any problems with the math videos either.
Thanks again for the math videos.

[VolksdeutscheSS]

Great exercise. However: I did this problem a little differently. I changed the bases of the logarithms twice:

1/log x 7 + 1/log x 8 = 1
log x 7 + log x 8 = (log x 7) (log x 8)
log x (7*8) = (log x 7) (log x 8)
ln (56)/ln x = (ln 7/ln x)(ln 8/ln x)
ln (56)/ln x = (ln 7)(ln 8)/[ln x]^2
ln (56) ln x = (ln 7)(ln 8)
ln x = (ln 7)(ln 8)/ln (56)
x = e^[(ln 7)(ln 8)/ln 56]
x approx. 2.7322
It's always great to get the same answer with different techniques. Sorry about the sloppy
nomenclature: I used Word as an editor. 🤣

[charlesmitchell]

Not Really an easy problem but you explained it clearly step by step.

[moeberry]

It’s a very easy equation there is no need for all these steps. Apply change of base formula immediately and factor out log(x) and solve in seconds.

[vhsy]

Loga(x) + Logb(x) =1

X = e^(Log(a) . Log(b) / [Log(a) + Log(b)])

Where Log(x) is the natural logarithm
Loga(x) is logarithm base a
e=2.71828...

[crmf]

Best math teacher on YouTube! Keep it, professor! You're awesome!

[brinzanalexandru]

I did the substitution X=7^k and we obtain k+k(log8(7))=1 from where easily k=1/(1+log8(7)) and X=7^(1/(1+log8(7))).

[oguzhanozdogan]

Cevabın farklı şekillerde ifade edilebileceğini söyleyebilirim.
logx*(1/(log7)+1/(log8)=1 =>
logx=1/(1/(log7)+1/(log8) =>
logx=(log8*log7)/(log56) =>
x=10^(log8*log7)/(log56)
Türkiyeden Selamlar .