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

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[@goldfing5898]

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...

[@flyingspirit3549]

Extra credit for going to all the work of showing how this is solved!

[@pierrecloutier9667]

Exactly 2,3219280948875

[@jeanlismonde8718]

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

[@broytingaravsol]

x=log_(2)(5)

[@orchestra2603]

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.

[@SyberMath]

We'll get to that

[@ytlongbeach]

@sybermath has there been a follow-up, on how to solve/simplify the log of the complex number solutions?

[@popitripodi573]

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

[@milandus1400]

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 🙂

[@kianmath71]

X= ln(5)/ln(2)

[@jankjaer1731]

This is the correct method.

[@francis-dt2hl]

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

[@AnirbanMukherjee-gv2we]

😂u can't write diff no. Both are having power x

[@ytlongbeach]

How does one solve the log of a complex number ?

[@SyberMath]

natural log which I call Ln (I wrote L in lowercase below) can be used with complex numbers as follows: ln(a + bi) = ln|a + bi| + i・arg(a + bi)

[@CrawliestCotter]

Also for imaginary numbers, as a result of Euler's Identity, ln(i) = i*pi/2. So in calculations of ln(a*i) you can simplify to ln(a) + ln(i) = ln(a) + i*pi/2.

[@akhileshiyer296]

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?

[@SyberMath]

it's not impossible. it just gives us complex solutions

[@akhileshiyer296]

@@SyberMath how can you take the natural log of a negative number?

[@EMERGINGGamer]

you can actually by converting complex number to eulers form @@akhileshiyer296

[@UttkarshPandey-u7p]

Aah u can se Euler form taking e as base ​@@akhileshiyer296

[@CrawliestCotter]

Logarithms with a negative domain have a complex range. You can calculate them using Euler's Identity e^(i*pi)=-1. From there you get i*pi=ln(-1). So ln(-x)= ln(x) + ln(-1) = ln(x)+ i*pi.

[@Sgt_Bill_T_Co]

2.322 as near as dammit. Done in 1 minute. first guess 2.2 then 2.3 then 2.32 - finaly 2.322 - peasy!

[@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

[@matematicacomailton]

beautiful bro...

[@SyberMath]

Thanks ✌️

[@AsifAsif-zy4gp]

Very good

[@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

[@luladrgn9155]

there are imaginary solutions, but u cant log them Lmao

[@msathwik8729]

​​@@luladrgn9155 you can log imaginary numbers bro. Like ln(i)=i(pi)/2

[@orchestra2603]

@@msathwik8729 i = exp(i * (pi/2 + 2pi*n) ), ln(i) = i*pi/2 + i*2pi*n, for any integer n=0,1,2... Complex logarithm doesn't give you a unique number, it's a "multi-branch" function, and i*pi/2 is only its principal value.

[@mathswan1607]

2^x=5 x=log_2(5)

[@Jefferytheturtlegod]

13

[@satyanarayana7346]

Log 5/2

[@ilyashick3178]

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.

[@maguschu]

Because the property of log is log(AB) = log A + log B but not log(A+B) = log A + log B. Thus, log_2(8^x + 2^x) cannot be separated as log_2 8^x + log_2 2^x.

[@ilyashick3178]

Each part of function not whole function a+b+c as log a+logb+;logc not log(a+b+c),

[@orchestra2603]

Well, that's because the equation with logarithms has nothing in common with the initial equation. I don't even understand why you constructed it. There's no logical link from one to the other. They are just two different things, and, no wonder, their are solutions are also different. So the roots that you found has nothing to do with the roots that are sought after.

[@ilyashick3178]

@@orchestra2603 Just in case, Log was introduced from 16 centure to get % of loan in math. Most of those lessons is nothing to do with college algebra or calculus.

[@rakenzarnsworld2]

125 + 5 = 130

[@Mpstories87]

X=ln_(2)(5)

[@goldfing5898]

You must write either log_2_(5) with low index 2 for base 2, or ln(5)/ln(2)

[@Mpstories87]

@@goldfing5898 definitely

[@devondevon4366]

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

[@robertveith6383]

No, it's 2^(3x) + 2^x - 130 = 0. You need the grouping symbols.

[@borisjeud8464]

X=2.32...

[@N00MAK_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