Finding the Real Solution | Exponential Equation | 5^x +5^3x = 10

#solvingexponentialequation #exponentialequation #howtosolveexponentialequation
In this video, we will solve this nice exponential 5^x +5^3x = 10. This is unusual exponential equation and we will find the real solution

[ministryoftruth]

5^x + 5^(3x) = 10 Re-write the right side as:
5^x +(5^x)^3 = 10 Let a = 5^x then:
a + a^3 = 10 Factor out a:
a(1 + a^2) = 10 Re-write 10 as 2( 1 + 4):
a(1 + a^2) = 2(1 + 4) 4 can be written as 2^2 so:
a(1 + a^2) = 2(1 + 2^2) Since both sides of the equation are in the same form, we can conclude:
a = 2 a = 2 = 5^x then:
5^x = 2 Taking the natural log of both sides, we have:
ln5^x = ln2 Which implies:
x ln5 = ln2 Which implies:
x = ln2/ln5 ≈ 0.4307
Thanks for the problem : )

[prabhudasmandal]

Factorisation of y^3+y--10=0 is easily determined. Since y=2 makes it 0, obviously (y--2) is one of the factors. So, y^2(y--2) +2y(y--2) +5(y--2) =0 or (y--2) (y^2+2y+5) =0.therafter follows the path as shown.

[Fred-yqfs]

Pretty simple.
Let X= 5^x, we have
X+X^3=10
obvious solution: X = 2
5^x = 2
xLn(5)=Ln(2)
x=Ln(2)/Ln(5)
f(X) = X^3+X-10
f'(x) = 3X^2+1 > 0
so f is always increasing, so f has only one real root: X = 2
So the unique solution is x = Ln(2)/Ln(5)

[shoshosalah]

It will be real at only one value where y=2
So
y=5^x
x=ln2/ln5~0.43

[DrLiangMath]

Nice question and wonderful explanation! 👍

[sie_khoentjoeng]

Anithe way, but not follow akgorithm of math, 😄😄😄

5^x + 5^3x = 10
5^x + (5^x) 3 = 10
a + a^3 = 10, if 5^x = a
a + a^3 = 2 + 8 = 2 + 2^3
Then a = 2 = 5^x
5^x = 2, or x = log 2 / log 5

[footballfdp]

4 line process has taken about 4 pages. But great explain

[renesperb]

It took me 1sec. to find the solution x = 0.