Solving An Interesting Log Equation

Solving An Interesting Log Equation | Math Olympiads

Рет қаралды 38,978

Күн бұрын

Пікірлер: 53

@spelunkerd Жыл бұрын

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 Жыл бұрын

💖 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 Жыл бұрын

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

@SyberMath Жыл бұрын

@@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 3 ай бұрын

@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 Жыл бұрын

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 Жыл бұрын

Agreed same

@SyberMath Жыл бұрын

Nice! 🤩

@MichaelRothwell1 Жыл бұрын

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 Жыл бұрын

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

@ytlongbeach Жыл бұрын

yup. i did it the same way !!

@erikdahlen9140 Жыл бұрын

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

@rajatdogra96 Жыл бұрын

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

@christianlopez1148 8 ай бұрын

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

@broytingaravsol Жыл бұрын

x=2^(ln3/ln6)

@KookyPiranha Жыл бұрын

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

@MushookieMan Жыл бұрын

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

@SyberMath Жыл бұрын

Very good!

@tetsuyaikeda4319 Жыл бұрын

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

@MarCamus Жыл бұрын

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

@SyberMath Жыл бұрын

Nice!

@somnathkundu7354 Жыл бұрын

Do the change of base and add the terms.

@SyberMath Жыл бұрын

Good thinking!

@Education-35 7 ай бұрын

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 🔥🔥🔥

@SyberMath 9 күн бұрын

Best of luck! 😍😍

@Education-35 9 күн бұрын

@@SyberMath very few months left for success 🔥🔥🔥

@moeberry8226 Жыл бұрын

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 Жыл бұрын

Nice! 🤩

@royfeer8651 Жыл бұрын

smart way!

@oahuhawaii2141 3 ай бұрын

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 3 ай бұрын

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

@josemanuelbarrenadevalenci653 Жыл бұрын

Confusedly explained

@MrGeorge1896 6 ай бұрын

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...

@giuseppemalaguti435 Жыл бұрын

x=3^log6(2)

@scottleung9587 Жыл бұрын

I got the same answer the first method had.

@popitripodi573 Жыл бұрын

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

@SyberMath Жыл бұрын

Great job! 🥰

@aproplayer8581 Жыл бұрын

I too used change of base method.

@MichaelRothwell1 Жыл бұрын

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).