Solving 8^x+2^x=130 | An Exponential Equation

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[goldfing]

8^x + 2^x = 130
(2*2*2)^x + 2^x = 130
2^x * 2^x * 2^x + 2^x = 130
Substitute 2^x = t
t*t*t + t = 130
t^3 + t = 130
t^3 + t - 130 = 0
This is a reduced cubic equation. Try to find an integer solution by testing the divisors of the constant 130.
1 * 130
2 * 65
5 * 26
10 * 13
Positive integers: t = 5 is a solution, since 5^3 + 5 = 125 + 5 = 130
There are no negative solutions because if t is negative, so is t^3 and the sum of them, so this cannot be positive 130.
Polynomial division:
(t^3 + t - 130) : (t - 5) = t^2 + 5t + 26
The remaining quadratic equation
t^2 + 5t + 26 = 0
as expected, has complex solutions only:
t2, 3 = (-5 +- sqrt(79)*i) / 2
= -2.5 +- 4.4441*i
(approximately).
I assume we are only interested in real solutions.
t = 5
2^x = 5
x = log2(5) = ld(5) = 2.321928095...

[jeanlismonde]

on pose u = 2^x et l'équation devient u^3 + u - 130 = 0 la factorisation est aisée avec (u - 5)(u²+5u +26)=0 donc une racine réelle en u soit u=5 et donc x = ln5/ln2
et deux racines complexes conjuguées en u : -5/2 + ou - i.rac(79)/2

[orchestra]

Nice! But how about talking in more details about a log of a complex number? It would be actually quite curious, since complex logarith, is a wider topic. In fact, you'll get an infinite number of solution for x over C.

[milandus]

Very nice!
My method was as follows:
once we have LHS in the form of (2^x)^3 + 2^x ... then let's rewrite our RHS to the same form,
RHS = a^3 + a
130 = a^3 + a
voilá 5^3 +5 = a^3 + a
a=5
comparing LHS = RHS
(2^x)^3 + 2^x = 5^3 +5
2^x = 5
x=lg(2) of 5 🙂

[popitripodi]

Very nice and pretty easy! I guessed that 5 was a root so I used Horner to find the other factor ❤❤❤❤

[francis-dthl]

8(2) + 2(6)=130 an English major who loves maths

[Elceszar]

Si ; 130 = 13×5×2
Entonces (8^X + 2^X) es multuplo de
13; 5 ó 2

Haciendo : 2^X = a
8^X + 2^X = 130 restando 2a
a^3 + a - 2a = 130 - 2a
a^3 - a = 130 - 2a
a(a^2 - 1) = 130 - 2a
a (a-1)(a+1) = (a-1)(a)(a+1) = 130 -2a

y como todos sabemos... el producto de
tres numeros consecutivos es multiplo de seis ( mod6) ó (°6)
mod6 = 130 - 2a
°6 = 130 -2a - 132 ; ojo 132 = °6
°6 = -2a -2 => °6 = 2a+2
°6 = 2(a+1) --> (a+1) = °6

Remplazando a = 2^X en (a+1)
2^X + 1 = 6

2^X = 5 ----> X = lg en base 2 de 5

No olvidemos que
2^X ( 2^2X + 1) = 130 = 13×5×2
Observar que es multiplo de 5

[akhileshiyer]

Do we not calculate if the complex solution is actually possible or not? That is, it the logarithm of that number a positive value not?
ln(r) + i(theta) has ln(-5) which is an impossible calculation, right?

[ilyashick]

Hmm, 130 is equal 128+2, 128 is 2^7, and total on right 2^7+ 2^1, take log on base 2 for each element of function log_2 of 8^x+log_2 of 2^x=log_2 of 2^7+log_2 of 2. 3x+x=8, x=2. But result 8^2 + 2^2 is not equal 130.

[barteqw]

Let t = 2^x
t^3 + t = 5^3 + 5
Then t^3 - 5^3 + t - 5 = 0
(t-5) (t^2+5t+26) = 0
t = 5
Δ < 0 so second bracket don't have any solutions
2^x = 5
log 2 both sides
x = log_2 5

[devondevon]

Answer Log 5 base 2 or 2.3219
2^3x + 2^x -130 =0 l
let n= 2^x . Hence
n^3 + n -130= 0
n^3- 125 +n-5 =0
(n-5)(n^2 + 5n +25) +1 ( n-5) = 0
n=5 is a solution
hence 2^x=5
x = log argument 5 , base 2 Answer

[NMAK_CA_Finalist]

Great but i guess this concept of "i" in the mathematics is quite illogical!!!
I mean how can you someone consider an imaginary term in the answer and call it an answer?
Its okay I don't criticize you for the answer but it is the mathematics oops illogical mathematics is the problem