An Exponential Equation With Radicals

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[@lempicka6737]

10^20=100^10 x=100

[@wtspman]

You could have taken the limit as x approaches 0, to get an idea of where the y-intercept lies in relation to the constant. You did all the other calculus pieces, and then hand-waived the last bit, saying 20ln10 is really big so it’s probably only going to intersect at 1 point.

[@ManjulaMathew-wb3zn]

Since 10 is involved you could have taken log
Sqrt(x)logx =20=10*2=10log100
One to one correspondence gives x=100
Or x^sqrt(x)=10^20=(10^2)^10=100^10
One to one correspondence gives x=100

[@michaelbaum6796]

Nice equation👍

[@KennethChile]

Using your old other techniques that you teached us:
x^x^½ = 10^20 || ^½
x^[½ * x^½] = [10^20]^½
[x^½] ^ [x^½] = 10^10, then
x^½ = 10
x= 100

[@SyberMath]

Good work!

[@barakathaider6333]

👍

[@georgepaidas1132]

That's a nice one equation. Keep solving 💪💪💪

[@SyberMath]

Thank you, I will

[@cav1928]

Ir hoy take decimal logaritmo (base 10)=log, then √x log(x) = 10*2, then √x =10 therefore x=100 AND log(100)=2

[@AngryEgg6942]

x^sqrt(x) = x^x^(1/2) = 10^20
Raise each side to the power of half:
(x^x^(1/2))^(1/2)=(10^20)^(1/2)
(a^m)^n = (a^(m*n)) = (a^n)^m => (x^(1/2))^(x^(1/2)) = 10^(20*1/2) = 10^10 =>
x^(1/2) = 10 /()^2
x=100

[@andrei_nazimov]

10^20=100^10=100^√100
x=100
😊

[@NirDagan]

sqrt(x) => x is non negative, and as x=0 is not a solution' we have x>0. Let t=sqrt(x), t>0. We get (t^2)^t = t^(2t) = 10^20 = 10^(2*10). Hence t =10 is a solution. Now f(t)=t^(2t) is an increasing function for t>0, hence this solution is unique

[@yakupbuyankara5903]

X=100

[@trojanleo123]

x = 100;
Used Lambert W

[@duytuanlengo5038]

Transfer the equation to √x^√x=10^10
x>0, set y=√x > 0,
so y^y = 10^10
ln(y^y) = ln(10^10)
y.lny = 10.ln10
e^lny.lny = 10.ln10
W(e^lny.lny) = W(10.ln10)
lny = W(10.ln10)
y = e^W(10.ln10)
Therefore x =e^[2.W(10.ln10)]
How do you know this solution equal to exactly 100 if we used Lambert as above? Would you like to show me your detail steps?

[@rakenzarnsworld2]

x = 100

[@user-wf3zv7ds3b]

y=sqrt(x) --> x=y^2 --> y^(2y)=10^20 --> y^y=10^10 ---> y=10, x=100

[@neuralwarp]

Just substitute
u = sqrt x

[@SyberMath]

Good thinking!

[@YannLeBihanFractals]

It is obvious that 10^20 = 100^10 , so x=100

[@suryanarayanamurtyyellajos9590]

So easy,x=100

[@rob876]

(√x)^(2√x) = 10^20
√x = 10
x = 100

[@giuseppemalaguti435]

√x^√x=10^10..√x=10..x=100

[@SidneiMV]

I've solved much more fast.
√x^√x = 10^10
so x = 100

[@SyberMath]

Wow! nice

[@mcumer]

If we take the sqare root of both members, we obtain: (sqrtx)^( sqrt x) = 10^10.. therefore sqrt x=10 and so x=100)... my idea is like @giuseppemalaguti435' s idea

[@spookbok]

((sqrt(x))^2)^(sqrt(x))=10^20=(10^2)^10
sqrt(x)=10
x=100

[@doichecabano]

√х=t; (t^2)^t=(10^2)^10; t=10; x=100