Solve the Logarithmic Equation with Different Bases | Math Olympiad Training

[@hansschotterradler3772]

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

[@PreMath]

No worries. We are all lifelong learners! I'll be sharing more videos on this topic shortly. Thank you for your feedback! Cheers! You are awesome, Hans 😀

[@Skank_and_Gutterboy]

Absolutely, these problems have helped me a lot with logarithms. I dealt with them in high school 30 years ago and not much since. I'm seeing them again in some of my engineering work and forays into chemistry. These online problems help a lot.

[@VolksdeutscheSS]

Hans: no one ever covered these types of problems in our school. High school only covered very basic--sehr grundsatzlicher Logarithmem.

[@hansschotterradler3772]

@@VolksdeutscheSS kann mich nich so gut dran erinnern. Ich glaube Logarithmen war vielleicht ein paar Wochen lang das Thema in der 10 Klasse (86/87 für mich).

[@VolksdeutscheSS]

@@hansschotterradler3772 Ja, Hans. Meine Lehrern haben uns ein bischen gelehrt uber Logarithmen aber bloB grundsatzliches Stoff.

[@justabunga1]

We can also use the change of base formula as log(x)/log(7)+log(x)/log(8)=1. 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 log(x)=1/(1/log(7)+1/log(8))=log(7)log(8)/log(56). Solving for x will be 10^(log(8^log(7))/log(56))=8^(log_56(7)) or 7^(log_56(8)).

[@Skank_and_Gutterboy]

This is essentially the route that I took. I used natural logs (not that it matters), got x=e^(ln(7)ln(8)/ln(56)), which simplifies to x=7^log(8, base 56). I did it 99% like you did, which sure seems more efficient and straightforward than the video method.

[@dreael]

@@Skank_and_Gutterboy I use exactly the same way as you. Formula =EXP(LN(7)*LN(8)/LN(56)) in Excel results in 2.732537123

[@PreMath]

Nice approach. Thank you for your feedback! Cheers! You are awesome, Justin. Keep it up 😀

[@mabdinur85]

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)))

[@christianderivi435]

I agree with you. I use log_7(x) in the same path like you. 👍

[@manueld848]

@@christianderivi435 I agree, but maybe it is better to convert both logarithms to decimals or naturals to easily get the result with the calculator at the end. Doing it like this I get, 2.7325... and no matter how much I review it, I can't find an error.

[@Mothuzad]

Yeah, this base change solution was what I did too, and it was so much easier that I could even do it without paper. I think it was at the very edge of my mental maths capabilities, so I found it really stimulating.

[@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.

[@anatoliy3323]

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

[@PreMath]

Dear Anatoliy, your comments are always a breath of fresh air! We are all lifelong learners! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Peace on Earth ☮

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@xyz9250]

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.

[@PreMath]

Many approaches possible! Thank you for your feedback! Cheers! You are awesome, XYZ 😀

[@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).

[@rayn1938]

Why if Im using a way from log8(x)+log7(x)=log7(7) , i am get 56^u = 7 not 8, something different in here , is true ?

[@ishaanlohani]

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

[@Mekenyejustus]

Great math🙏🙏 Alternatively (Logx/log7)+(logx/log8)=1 logx(1/log7+1/log8)=1 logx{(log7+log8)/(log7log8)}=1 logx=(log7log8)/(log7+log8) x=10^{(log7log8)/(log7+log8)} ~2.733

[@justicechristopher9780]

Please, if you don't mind; why should the logs be equated to the same constant, since both logs are clearly different and even having different bases?

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@ronellmonieno4353]

Math professor is always calm. What country are you from siR?

[@charlesmitchell5841]

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

[@devondevon4366]

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

[@PreMath]

Wow, Well done! Thank you for sharing! Cheers! 😀 You are awesome, Devon. Keep it up 👍

[@hanswust6972]

I feel so good knowing there is a solution!

[@FaixaPretanoENEM]

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

[@PreMath]

Wow, thanks! So nice of you. Cheers! You are the best, CRMF 😀 Love and prayers from the USA!

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@ruilongsheng2845]

(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

[@whaddoiknow6519]

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.

[@Mathskylive]

Về cơ bản chúng ta đưa về một cơ số. Bài toán hay. Cảm ơn bạn nhé.

[@vh73sy]

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...

[@appybane8481]

they can be simplified as a^(log_(ab) a)

[@vh73sy]

@@appybane8481 Assuming a & b are positive It simplifies to a ^ ( logb / log(ab))

[@gokublue3059]

Obrigado. Brazil aqui!

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@christianderivi435]

Very nice! Thanks 👍

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@josueregaladolopez3202]

Well done! I liked your procedure

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@studytotest4839]

whch clas question

[@oguzhanozdogan4915]

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 .

[@brinzanalexandru2150]

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))).

[@PreMath]

Thank you for sharing! Cheers! 😀 You are awesome, Brinzan. Keep it up 👍

[@prabhudasmandal6429]

Nice problem is nicely solved thanks.

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@davidfromstow]

All went completely over my head but I still enjoyed it!

[@PreMath]

No worries. We are all lifelong learners! I like your positive attitude😀 Thank you for your feedback! Cheers! You are awesome, David 😀 Love and prayers from the USA!

[@MdFahim-of7hc]

nice math.Your videos are always good.

[@PreMath]

Glad you think so! Thank you! Cheers! 😀 You are awesome, Mr. Legend. Keep it up 👍

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@HappyFamilyOnline]

Nicely explained 👍 Keep sharing😊

[@PreMath]

Thanks a lot 😊 You are awesome 😀

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@jesusantoniocarhuashuerta4662]

Excellent.

[@moeberry8226]

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.

[@ronaldnoll3247]

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

[@HaiderBB-gq3li]

I solve it and x =√7 and when I check it gonna be nearly 1

[@broytingaravsol]

the focuses of bases deviate

[@laggyluke5700]

8^log56(7) is a proper answer too. ;)

[@justabunga1]

That answer works too. There is a way to get that answer also.

[@PreMath]

Yes! Thank you! Cheers! 😀 You are awesome, Luke. Keep it up 👍

[@nicogehren6566]

very nice question

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome, Nico 😀

[@sandanadurair5862]

I enjoyed the log manipulations.

[@PreMath]

Glad you think so! Thank you! Cheers! 😀 You are awesome, Sandanadurai. Keep it up 👍

[@242math]

very well explained, thanks for sharing

[@PreMath]

You are very welcome. So nice of you. Thank you for your feedback! Cheers! You are awesome 😀

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@mengpheapMEF]

Thanks

[@giuseppemalaguti435]

7^(1/(1+log7)),con log= logaritmo in base 8

[@rahulpaul5539]

Great approach

[@PreMath]

Glad you think so! Thank you! Cheers! 😀 You are awesome, Rahul. Keep it up 👍

[@chandrashekharmehta6121]

My answer was 343 to the power log 2 with base 56 Which I then realise it is the correct answer..

[@mahalakshmiganapathy6455]

Admirable,👌

[@PreMath]

Glad you think so! Thank you! Cheers! 😀 You are awesome, Mahalakshmi. Keep it up 👍

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@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. 🤣

[@twocsies]

This seems to be the more straightforward solution.

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome. Thank you! Cheers! 😀 You are awesome, Pranav. Keep it up 👍

[@gelbkehlchen]

Solution: log7(x)+log8(x) = 1 |After loga(x)=logc(x)*loga(c)=change of base, with a=7 and c=8: log7(x)=log8(x)*log7(8) ⟹ log8(x)*log7(8)+log8(x) = 1 ⟹ log8(x)*[log7(8)+1] = 1 |/[log7(8)+1] ⟹ log8(x) = 1/[log7(8)+1] ⟹ x = 8^{1/[log7(8)+1]} = 8^{1/[ln(8)/ln(7)+1]} = 2.732537123052263 Sample: left side: log7(2.732537123052263)+log8(2.732537123052263) = ln(2.732537123052263)/ln(7)+ln(2.732537123052263)/ln(8) = 1 right side: 1 Left side = right side, everything ok.

[@susennath6035]

Nice

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome, Susen 😀

[@SuperYoonHo]

great

[@nirupamasingh2948]

V nice way of solving

[@PreMath]

Glad you think so! Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@ryanchin5232]

Log change of base: to 56 base of log. few steps of simplifications you can get x=56^(log 56 base of 7 * log 56 base of 8) = 7^(Log 56 base of 8)

[@lavoiedereussite922]

thank

[@domenicosalsano9337]

L'ho risolto più facilmente cambiando subito la base al log 7 di x in log 8 di x / log 8 di 7..

[@lazaremoanang3116]

Simple, x=(ln7ln8)/(ln7+ln8).

[@Soumik78]

sir, more problem on logarithm

[@PreMath]

More interesting problems are on their way. Please stay tuned. Thank you for your feedback! Cheers! You are awesome, Soumik. Keep it up 😀

[@domingosmanuel9376]

kzbin.info/www/bejne/nn-nm3Wmd9qgoJo&ab_channel=BECOMEAGENIUS try also this way of solving you may find interesting

[@marklevin3236]

Much simpler Shift to same base and we will have a linear equation involving logarithms....

[@rmela4501]

I got e^[ln(7)*ln(8)/ln(56)]

[@azizurrehmanbaig8030]

Your box-writings disturb visibility of solutions- writings

[@MrTejomani]

5^x+7^y=174 X+Y=5 X=? Y=?

[@FrogworfKnight]

This was done before watching the video log7(x)+log8(x)=1 log(x)/log(7)+log(x)/log(8)=1 (log(8)log(x)+log(7)log(x))/(log(7)log(8))=1 log(x)(log(8)+log(7))/(log(7)log(8))=1 log(x)(log(56))/(log(7)log(8))=1 log(x)=(log(7)log(8))/log(56) log(x)=log(8^log(7))/log(56) log(x)=log(8^(log(7)/log(56))) x=8^(log(7)/log(56)) x=8^(log56(7)) After watching the video I was able to confirm that its the same value as the one presented in the video, which makes sense, since at the 6th line of work, log(8) and log(7) could have easily gone the other way around.

[@PreMath]

Very well done Thank you for your feedback! Cheers! You are awesome . Keep it up 😀

[@wanmuhammadhaiqealwan7034]

How do you know to convert log 7 log 8 to another form, such as log (8log7)? is a question I have. Can you provide step-by-step instructions?

[@FrogworfKnight]

@@wanmuhammadhaiqealwan7034 I think I need context of what you mean. I had thought you meant a change base formula, but there are several kinds of form changes I did in this problem along side the change base formula, namely the use of the product rule and the exponent rule. As a reminder, the change base formula is logA(b)=log(b)/log(A), the product rule of logarithms is log(ab)=log(a)+log(b), and the exponent rule goes log(a^x)=xlog(a). There is no point however where I wrote in log(8log(7)) so I don't know which example you might be referring to. The closest I can find is on lines 5 and 6 where the exponent rule was applied thusly: log(7)log(8)=log(8^log(7))

[@wanmuhammadhaiqealwan7034]

@@FrogworfKnight Actually, you have answered my question clearly, and I am grateful for your reply; thank you. I just tried again to solve the question using your answer from line 5 and changed the base and power rules, and I got this: log(x) = (log(7) * log(8)) / (log(56)) log(x) = (log8(8)/log8(10))* (log(7)/log(56)) log(x) = (log8(8)/log8(10))* (log56(7)) log(x) = ((log8(8))(log56(7))/log8(10)) log(x) = (log8(8^log56(7))/log8(10)) log(x) = log(8^log56(7)

[@FrogworfKnight]

@@wanmuhammadhaiqealwan7034 You are just one step away from matching my answer then.

[@davidmcknlght2700]

👍

[@AllanKobelansky]

Ah. My daily dose of math problems.

[@PreMath]

Glad to hear that! Thank you! Cheers! 😀 You are awesome, Allan. Keep it up 👍

[@paolophoenix]

Change the base!.

[@giuseppemalaguti435]

Ciao

[@nezhachannel2943]

👉👍

[@PreMath]

Thank you. Cheers! You are awesome, Nezha 😀

[@mariangorski]

:O :O

[@mclarenforever7798]

Your method is too long, I solved in just 4 lines.

[@dicksonphisthur3398]

Yuk. Very messy.

[@devondevon4366]

I completely different approach. Since 7 and 8 are raised by two numbers (let one number = p the other m) 7 and 8 are raised to (or the same argument 'x') as (p+ 1-p=1 hence 7^1-p = x AND 8^p =x hence 7^p-1= 8^p log 7-log7p= log 8p log 7= log 56p (7x8) log7/log56 =p hence x= 8^ log 7/log56 or 2.7325

[@kandakoeipudsa5014]

Thanks