An Exponential Rational Equation

[@pizza8725]

The imaginary solutions are (7ln(2)+2ipin)/ln(2) You did the solution when n=0 which is real

[@TejasDhuri-p8z]

Let 2^x=m, m³-1/m²+m+1=127 a³-b³=(a-b)(a²+ab+b²) (m-1)(m²+m+1)/(m²+m+1)=127 m²+m+1 !=0, Thus m-1=127 m=128 2^x=2⁷ Thus x=7

[@arekkrolak6320]

Technically you dont have to depress cubic equation to solve it, we usually do it because then the formulas are much simpler, but there is also formula for cubic in standard form

[@SyberMath]

what is it? can you share?

[@gaborpaul]

Most straightforward: Add 1 to both sides of the equation, obtaining RHS=2^7. On the LHS, adding one is equivalent to a fraction in which the new numerator is the sum of the old numerator and the denominator. This means that the constants (-1 and +1) vanish, and the new numerator is simply 2^x times the denominator. That means that the equation leads to 2^x = 2^7, i.e., x=7.

[@kailashanand5086]

That was crazy!!! Nice find

[@vaggelissmyrniotis2194]

X=7+2kπi/ln2 the complex solution.

[@neuralwarp]

But you ^didn't^ show us the first method. You rambled about it but didn't show it.

[@KyyTyy]

😂😂😂

[@SyberMath]

left as an exercise for you 😄

[@nirajkumarverma5299]

No complex solutions. If considering complex roots, by cross multiplication, it will make the Numerator and denominator zero. Which is not acceptable.

[@MathNerd1729]

There actually are complex solutions for x since 2^x = 128 has complex solutions other than x = 7. For example, x = 7 + 2πi/Ln(2) is also a solution to the equation.

[@MsAcpaul]

​@@MathNerd1729He is right actually. The bounds of the solution means that 4^x + 2^x + 1 ≠ 0 as a denominator cannot be zero.

[@MathNerd1729]

@@MsAcpaul While I agree that the denominator cannot equal 0, I was trying to point out that once you get 2^x = 128, there are still other complex numbers that satisfy that equation other than x = 7

[@SyberMath]

Agreed! you can write the general solution as x = 7 + i[(2πn)/ln(2)]

[@E.h.a.b]

Let 2^x = y y^3 - 1 = 127(y^2 + y + 1) (y - 1)(y^2 + y + 1) = 127(y^2 + y + 1) (y^2 + y + 1) (y-1 -127) = 0 (y^2 + y + 1) (y - 128) = 0 y^2 + y + 1 = 0 OR y - 128=0 No real root OR y = 128 2^x = 128 = 2^7 x = 7

[@scottleung9587]

I also got x=7 as the only real solution. There are no complex solutions, since the quadratic cannot be zero after simplifying the fraction. Who am I kidding, I solved for them anyway!

[@MathNerd1729]

There actually are complex solutions for x since 2^x = 128 has complex solutions other than x = 7. For example, x = 7 + 2πi/Ln(2) is also a solution to the equation.

[@Hemicubesquare]

x=7

[@giuseppemalaguti435]

2^x-1=127...x=7

[@AntonioOlivares-on3fl]

x = 7

[@rakenzarnsworld2]

x = 7