# Solve in one Minute | Learn how to solve exponential equation quickly | Math Olympiad Training

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This is so cool Professor!

This one is quite unique in my opinion

Well explained sir

V nice explaining the ruleof indicesv clearly

Thnku

Wow this one was too easy

5th root 10

let y=x^5, get x = y^(1/5)

-> (y^(1/5)^y = 10^2

-> y* lg(y^(1/5)) = 2

-> y * lg(y) = 10

-> y^y = 10^10

-> y = 10

-> x = y^(1/5) = 10^(1/5)

Nice video sir

Nice method

i made the substitution u=x^5 so

u^1/5u=100

u^u=100^5=10^10

u=10

x=5th root of 10

There is an issue:

a^b^c = a^(b^c) or a^b^c = (a^b)^c,

in general case a^(b^c) is not equal to (a^b)^c.

So, I think we should clarify this issue before solve the problem……

Nice and easy question

a^a=b^b———>a=b ,if and b ,a>=1 or if 0.368<a,b<=1

Thanks for video. Good luck sir!!!!!!!

nice.

remember my JEE advance preparetion time.

nostralgic

https://youtu.be/1Jp6IcxM8j4 this is new math channel

Outstanding explain

a^a =b^b implies that a=b? Is that always true? A graph of y=x^x curves back on itself near zero so has repeat values so I would say this is not generally true. Anyone?

Siuuuuu, thanks for shearing

I have learn another word!! Xd

I used Lambert W function to solve this: x^5 ×lnx=ln100 —》x^5 × ln(x^5)=ln( 10^10)—》W(ln(x^5) × e^(ln(x^5)))=W(10ln10)—》ln10=ln(x^5)—》x=10^0.2

I got as far as X*LOG(X)=2/5 before the crystal ball went dark. I had no clue. Very clever solution. Thanks!

very well done, thanks for sharing this Math Olympiad question

I like multi-decker exponents. It looks so nice<)) Thank you so much, Mr PreMath! Stay blessed and all the best to you!

Beautiful solution

Excellent sir

Thanks sir.love from Bangladesh

Herr professor, I saw your solution and I didn't understand it, as I believe that the value of X = 1.7124

How I made: X^X = Y

2) Y^5 = 100

Y = 2,5118

2,5118 = X^X

X = 1,7124

where did I go wrong?

Well, I got that completely wrong – most interesting and I thank you!

amazing …https://youtu.be/77W9zhTBuUM

x^x^5=100 add 100 to both sides

x^x^5=10^2 convert 100 to exponent form

(x^x^5)^5= (10^2)^5 raised both sides to the power of 5

x^5^x^5=10^10 now there is a pattern, x^5 's base matches 10 (base), and x^5's exponent matches 10 (exponent) hence

x^5=10

x= 10^1/5 or x=10^0.2

I wish the video had discussed the secondary root of 100, so that when 100 was rewritten as 10^2 we also looked at (-10)^2

✅

Manipulation at its best!

x=10^1/5 or 10^0.2 or 1.5849

fifth root of 10

Very well done. I like your technique. However, I got the same answer you did by putting the equation into the "Lambert form":

ln x^(x^5) = ln 100

x^5 [ln x] = 4.605

e^ [ln (x^5)] * ln x = 4.605

e^ [5 ln (x)] * lnx = 4.605

5 ( e^ [5 ln (x)] * lnx) = 5* 4.605

( e^ [5 ln (x)] * 5 lnx) = 23.025

Lambert both sides and get. . .

5 lnx = productlog (0, 23.025) from Wolfram Alpha

5 ln x = 2.30256

ln x = 0.460512

e^(ln x) = e^0.460512 is approx. 1.585 which also equals approx. 10^1/5

x = 1.585.

Nice to get the same answer with two different techniques.

Only if you test the solution

x= 10^(1/5)=1.58489319

then you get with

(x^x) = 2.07481146

and

(x^x)^5= 38.44972 and NOT 100!

Where am I wrong?

@PreMath I am really proud to be on your channel, I am an electrical engineer,47 yrs old & I have liked Math since I was a child, when I watch your videos & try solving before seeing the solution, it's a great pleasure, so many thanks for you

Can u solve it with log?

There is a easier way to do it

X^5=t

So X= t^1/5

X^X^5=100

Log base 10 on both sides

X^5logX=2log10

t/5logt=2

tlogt=10

t^t=10^10( because of log base 10)

t=10

So X= t^1/5=10^1/5

If y is substituted for x^5 the equation becomes y^(y/5) = 10^2 and the solution is obvious.

Like

ha ha ha….

Shouldnt this problem have 5 roots?

Let X =2y

好

好，奖 1 万元

But what is 10 to the 1 fifth?

By inspection, x==1 is too small. X==2 gives 2^32 which is too big. So 1 < x < 2. Trying the mean value x == 3/2 looked close. I could not be bothered doing any more than that.