Solving An Interesting Log Equation | Math Olympiads

[@spelunkerd]

What I like about your videos is how you consistently do not choose the obvious pathway that most of us would use. In this case rather than following the blaring siren to change base, you chose more elegant pathways to get the same place. Just because they may take a little longer at times does not take away the theme that there are often many options. Along the way we recall principles that aren't necessarily used every day.

[@SyberMath]

💖 Thank you for the feedback! My mindset seems to be "How can I make the solution more complicated?" It's funny that I haven't thought of the "change of base" method for this problem. Thanks for the heads up!!!

[@leif1075]

@@SyberMath Thank you for sharing. Wouldn't you agree your teo methods here are the same method thoigh just swapping bases?

[@SyberMath]

@@leif1075 no problem! That’s right. They are kind of the same. Using change of base, which, interestingly I haven’t thought of btw, would be an entirely different approach.

[@oahuhawaii2141]

@SyberMath: I did the change to a common base, and got a general form; then, I could choose base e, 2, 3, 10, ... log2(x) + log3(x) = 1, switch to a common base log(x)/log(2) + log(x)/log(3) = 1 log(x)*(log(3) + log(2)) = log(2)*log(3) log(x) = log(2)*log(3)/log(6), choose base e x = e^[ln(2)*ln(3)/ln(6)] = e^(1/(1/ln(2) + 1/ln(3))) ≈ 1.5295923284918... BTW, with the exponent having a product and a quotient of logs, we can change bases for exponentiation and logarithm. Arranging for e^ln(2): x = e^[ln(2)*ln(3)/ln(6)] = [e^ln(2)]^[ln(3)/ln(6)] = 2^log6(3) This is the same if we chose base 2 instead of e. Arranging for e^ln(3): x = e^[ln(2)*ln(3)/ln(6)] = [e^ln(3)]^[ln(2)/ln(6)] = 3^log6(2) This is the same if we chose base 3 instead of e.

[@justabunga1]

There is another method that you can also use, which is the change of base formula. This equation can be rewritten as log(x)/log(2)+log(x)/log(3)=1. Multiply everything by the LCD, which is log(2)log(3), and factor out log(x). This means that (log(3)+log(2))log(x)=log(2)log(3). Divide both sides by log(3)+log(2), or log(6). log(x)=log(2)log(3)/log(6), and x=10^(log(2)log(3)/log(6)). Simplify this even better and using properties of logarithms, which is x=2^(log_6(3)). The answer can also be x=3^(log_6(2)) if we interchange the the base of the exponent and the log of an argument.

[@rajatdogra96]

Agreed same

[@SyberMath]

Nice! 🤩

[@MichaelRothwell1]

Agreed, that was my method (though I used ln rather than log). I was very surprised that SyberMath didn't present this method also, as the equation is crying out for the change of base rule! Also, this method leads to a pleasing symmetrical result, easily simplified to the forms found in the video.

[@iz8376]

yeah I did this method in my olympiad too and saves a lot of time

[@ytlongbeach]

yup. i did it the same way !!

[@erikdahlen9140]

I did log base change in my head when i saw the problem, factored, and solved for x. easier than whatever you did here

[@MrGeorge1896]

I suggest using ln(b) / ln(a) instead of log_a(b). ln(x) (1 / ln(2) +1 / ln(3)) = 1 ln(x) = ln(2) * ln(3) / ln(6) so x = e ^ (ln(2) * ln(3) / ln(6)) = 1.52959...

[@MushookieMan]

I just used the change of base formula on one of the logs, wasn't a bad method

[@SyberMath]

Very good!

[@MarCamus]

Changed both bases to ln and after some algebra I got the solution in terms of ln and e

[@SyberMath]

Nice!

[@rajatdogra96]

Log x ( log 2 + log 3) = log 2 * log 3

[@broytingaravsol]

x=2^(ln3/ln6)

[@christianlopez1148]

log2x=lnx/ln2, log3x=lnx/ln3....easyly you can find lnx.

[@somnathkundu7354]

Do the change of base and add the terms.

[@SyberMath]

Good thinking!

[@KookyPiranha]

Aint no way this is a math olympiad problem. I saw a similar one in my hw

[@Education-35]

My answer is same but another approach First i use reciprocal property To make same base And then take LCM And apply log(a) + log(b) = log(ab) After that Use base changing property And I got answer 2 JEE aspirant 2025 🔥🔥🔥

[@GurpreetSingh-PI]

ELEGANT - spy x family

[@moeberry8226]

The fastest way is to use change of bases. So you obtain log(x)/2+log(x)/3. Now just factor out a log(x) and divide by that constant on both sides and then anti log both sides and your done.

[@SyberMath]

Nice! 🤩

[@royfeer8651]

smart way!

[@oahuhawaii2141]

Not correct. The 2 should be log(2), and the 3 should be log(3). The base of the log can be chosen as 2, 3, e, 10, ...

[@moeberry8226]

@@oahuhawaii2141 your 100 percent right bro. That’s a typo on my end.

[@popitripodi573]

I used the change of base method!!your approach was very good alternative!!!!❤❤❤❤

[@SyberMath]

Great job! 🥰

[@aproplayer8581]

I too used change of base method.

[@josemanuelbarrenadevalenci653]

Confusedly explained

[@tetsuyaikeda4319]

I could know 'Zee is not see, nor c, nor be ' that must be some kind of joking

[@scottleung9587]

I got the same answer the first method had.

[@giuseppemalaguti435]

x=3^log6(2)

[@GreenMeansGOF]

change of base is how I would have done it

[@SyberMath]

Great!

[@rakenzarnsworld2]

x = 1.5

[@phongvong8639]

It is over I mean they are. Then something has to be prioritised. Please. Thank you.

[@jfcrow1]

x equal cube root of 3

[@gatosimple2354]

Profe menos palabras demás,sea más directo

[@MichaelRothwell1]

In case anyone is wondering, the reason for the rule a^(log꜀b)=b^(log꜀a) is that if you log each side to base c and use the power rule for logs, both sides give log꜀a log꜀b. Putting it another way, c^(log꜀a log꜀b)=(c^log꜀a)^log꜀b=a^log꜀b and c^(log꜀b log꜀a)=(c^log꜀b)^log꜀a=b^log꜀a so a^(log꜀b)=b^(log꜀a).

[@angelamusiemangela]

x=2-1