A Homemade Exponential Equation

[@prakrit1280]

Since powers of 3 are 1 and -1 [as 10/3 is 3^1 + 3^(-1) ], so 1 + (-1) = 0, that's (x+1) + (3x-9) = 0, or 4x - 8 = 0, or x = 2 done😊

[@marcgriselhubert3915]

The given equation is equivalent to 3^(((2.x -4)/(x -5)).(3^(-1) + 3^(1)) = 10/3, or 3^((2.x - 4)/(x - 5)) = 1, or (2.x -4)/(x -5) = 0, or x = 2. (Quick!)

[@erikroberts8307]

Excellent!!! 😊😊😊

[@SyberMath]

Thank you! 😃

[@user-lu6yg3vk9z]

@@SyberMath hey quick question what is the integral of inverse tangent going from -infinity to infinity?

[@bkkboy-cm3eb]

3^((x+1)/(x-5)+3^((3x-9)/(x-5) = 3^(1+6/(x-5))+3^(3+6/(x-5)) = 30·3^(6/(x-5))=10/3 → 3^(6/(x-5))=1/9 → 6/(x-5)=-2 → x-5=-3 ∴ x=2

[@AmirgabYT2185]

I did it the same way 😁

[@pwmiles56]

It seemed obvious to substitute u=x-5. x=u+5, x+1=u+6, 3x-9=3u+6 3^(1+6/u) + 3^(3+6/u) = 10/3 (3+3^3) 3^(6/u) = 10/3 30 3^(6/u) = 10/3 3^(6/u) = 10/90 = 1/9 6/u = -2 u = -3 x = 2

[@SyberMath]

Wow!

[@vaggelissmyrniotis2194]

Substituted with t=3^(1/(x-5)) and found x=2.

[@giuseppemalaguti435]

.....3*3^(6/x-5)+27*3^(6/x-5)=10/3...t=3^6/x-5..30t=10/3...t=1/9...6/x-5=-2...x=2

[@yoav613]

Nice problem that i solved in my head😊

[@SyberMath]

😮😁

[@SanePerson1]

I guess I just got lucky: I factored 3^[(x + 1)/(x - 5)] out of both terms on the lhs as the first step: 3^[(x + 1)/(x - 5)](1 + 3^[(2x - 10)/(x - 5)] = 10/3 - that seemed like the way to go because the denominators of both exponents are the same. Then everything falls out easily since (2x - 10)/(x - 5) = 2 ⇒ 3^[(x + 1)/(x - 5)](1 + 3²)] = 10/3. Obviously, 1 + 3² is just 10, se can divide both sides by 10: 3^[(x + 1)/(x - 5)] = 10/30 = ⅓ = 3⁻¹. We can set the exponents equal then, (x + 1)/(x - 5) = -1, which has the solution x = 2. Another 1 minute solution.

[@rakenzarnsworld2]

x = 2

[@scottleung9587]

I also got x=2 as the only solution.

[@Psykolord1989]

Before watching: Alright, in this case, I'm starting with the right hand side. I'm going to look for powers of 3 such that 3^a + 3^b = 10/3. Well, 10/3 = 9/3 + 1/3 = 3 + 3^(-1). So 1 and -1 are the powers of our two terms, in some order. Let's find if there are any Solution 1: (x+1)/(x-5) = -1, (3x-9)/(x-5)=1 (x+1)/(x-5)=-1 -> x+1 = -x+5 -> 2x=4 -> X=2. Checking this with the second term gives us 3^(-3/-3) = 3^1. Solution 2: (x+1)/(x-5) = 1, (3x-9)/(x-5)=-1 This does not give us a suitable value for X. (x+1)/(x-5) = 1 -> x+1 = x-5 -> 6=0. Thus, we discard solution 2. We are left with solution 1, X=2.

[@almanduku9043]

3 ¹⁺⁽⁶/ˣ⁻⁵⁾ + 3 ³⁺⁽⁶/ˣ⁻⁵⁾ = 10 . 3⁻¹ x - 5 = y => ..... x=2

[@GsBabu-sk6iv]

X = 7/2, is also a solution

[@howardhtml]

why is 7/2 a solution??? 3^(-3) + 3^(-1) is not equal to 10/3.

[@GsBabu-sk6iv]

@@howardhtml you are ✅correct

[@ibrahimmassy2753]

Another way: 3^((x+1)/(x-5))+3^((3x-9)/(x-5))=10/3 3^(1+6/(x-5))+3^(3+6/(x-5))=10/3 3(3^(6/(x-5)))+27(3^(6/(x-5)))=10/3 30(3^(6/(x-5)))=10/3 3^(6/(x-5))=1/9 6/(x-5)=-2 x=2

[@Quest3669]

Its shud be like 3+(1/ 3) now easily find in seconds 2 is the solln by all rhe way only

[@broytingaravsol]

the pattern

[@ViaanHossain]

Lol😂 homemade equation

[@drggayathridevi195]

Hi

[@devondevon4366]

x= 2 3* ([3^x+1]/[x-5]) + 3^]x -9]/[x-5]) = 10 multiply both sides by 3 3^x +2/x-5 + 3^x -8/x-5 =3^2 + 3^0 (as 10 can be expressed as 9 + 1 and 3^0 =1) (x+2)/(x-5) = 2/1 and (x+2)/(x-5)= 0 x+2=2x-10 x+2= (x-5)*0 x= 12 x=-2 (x-8)/(x-5) = 2/1 and (x-8)/(x-5) =0 x-8 = 2x+10 x-8=0 x=2 x=8

[@devondevon4366]

only when x =2, other numbers, - 2, 8 and 12 did not

[@jarikosonen4079]

Finding the right substitution seemed like long process...

[@SyberMath]

Check @pwmiles56's comment! Great ideas

[@VictorPensioner]

(1) 3^[(x + 1) /(x - 5)] + 3^[(3x - 9) /(x - 5)] = 10/3 Divide (1) by 3^[(x + 1) /(x - 5)]: 1 + 3^[(3x - 9) /(x - 5) - (x + 1) /(x - 5)] = (10/3) * 3^[-(x + 1) /(x - 5)] or 1 + 3^[(3x - 9 - x - 1) / (x - 5)] = 10 * 3^[-(x + 1) /(x - 5) - 1] or 1 + 3^[(2x - 10) / (x - 5)] = 10*3^[-(x + 1 + x - 5)/(x - 5)] or 1 + 3^[2(x - 5)/(x - 5) = 10 * 3^[-(2x - 4)/(x - 5)] or 1 + 3² = 10 * 3^[-(2x - 4)/(x - 5)] or 3^[-(2x - 4)/(x - 5)] = 1 => (2x - 4) = 0 => x = 2

[@TheLukeLsd]

Let x-5=k 3^(k+6)/k +3^(3k+6)/k=10/3. --> For k≠0 -> x≠5, 3^(k+6)/k×(1+3^2)=10/3 --> 3^(k+6)/k×10=10/3 --> 3^(k+6)/k=1/3 --> 3^(k+6)/k=3^-1 --> K+6/k=-1 --> 2k=-6 --> k=-3 --> x=2. c.q.d.

[@key_board_x]

3^[(x + 1)/(x - 5)] + 3^[(3x - 9)/(x - 5)] = 10/3 → where: x ≠ 5 3^[(x + 1)/(x - 5)] + 3^[(x + 2x - 10 + 1)/(x - 5)] = 10/3 3^[(x + 1)/(x - 5)] + 3^[(x + 1 + 2x - 10)/(x - 5)] = 10/3 3^[(x + 1)/(x - 5)] + 3^[{(x + 1) + (2x - 10)}/(x - 5)] = 10/3 3^[(x + 1)/(x - 5)] + 3^[{(x + 1) + 2.(x - 5)}/(x - 5)] = 10/3 3^[(x + 1)/(x - 5)] + 3^[{(x + 1)/(x - 5)} + {2.(x - 5)/(x - 5)}] = 10/3 → as x ≠ 5, you can simplify by (x - 5) 3^[(x + 1)/(x - 5)] + 3^[{(x + 1)/(x - 5)} + {2}] = 10/3 → let: a = (x + 1)/(x - 5) 3^(a) + 3^[a + 2] = 10/3 3^(a) + [3^(a) * 3^(2)] = 10/3 3^(a) + [9 * 3^(a)] = 10/3 3^(a) * 10 = 10/3 → you can simplify by 10 both sides 3^(a) = 1/3 3^(a) = 1/3^(1) 3^(a) = 3^(- 1) a = - 1 → recall: a = (x + 1)/(x - 5) (x + 1)/(x - 5) = - 1 (x + 1) = - (x - 5) x + 1 = - x + 5 2x = 4 → x = 2

[@trojanleo123]

x = 2