KCC's Quizzes AQQ253 about a funny equation

Answer is x= log₂ 5 ≈ 2.32193

(2^3)^x + 2^x =130

(2^x)^3 + 2^x = 130

put y= 2^x --> y^3 + y = 130, which is solved only by y=5, so 2^x = 5 and then x= log₂ 5

3 values for x. (log can be of any base)

a) log(5)/log(2)

b) log((-5 + i*sqrt(79))/2) / log(2)

c) log ((-5 - i*sqrt(79))/2) / log(2)

8^x + 2^x = 130, can be changed to 2^(3x) + 2^x = 130

Substitute for m = 2^x, rearrange. 

m^3 + m - 130 = 0

(m-5)(m^2 + 5m + 26) = 0

3 values of m,

m = 5, m = (-5 + (i*sqrt(79))) / 2, m = (-5 - (i*sqrt(79))) / 2

Substitute it values back to m = 2^x. Apply log of any base.

log(m)/log(2) = x

8^x+2^x=130

(2³)^x+2^x=130

(2^x)³+2^x=130

Let a=2^x which results in the cubic equation:

a³+a-130=0

By inspection, the real root is a=5.  The two other roots are non-real complex conjugate numbers.  See Cardano’s formula in Wikipedia for the calculation of the complex roots.

X=log(a)/log(2)=2.321928095

X=2.321928

Fully correct @NotBeans! Congratulations and big applause!

Correct! That equation accepts 3 solutions: one real and 2 complex. Congratulations

Bravo Roland,

x=2.321928 is well a solution to the equation, but it is not the only one!

May be you can find them?

8^x + 2^x = 130

Assume 2^x = t then the equation will become t^3+t-130 = 0

So possible value of t will be 5 so then equation will be (t - 5) (t^2+5t+26) = 0.

And t^2+5t+26 will give us imaginary roots because discrimant is 5*5-4*1*26 which is negative so possible value of t is 5 

Hence value of x is log5/log2 which is 2.3219

Ans 2.3219

8^x + 2^x = 130

Assume 2^x = t then the equation will become t^3+t-130 = 0

So possible value of t will be 5 so then equation will be (t - 5) (t^2+5t+26) = 0.

And t^2+5t+26 will give us imaginary roots because discrimant is 5*5-4*1*26 which is negative so possible value of t is 5 

Hence value of x is log5/log2 which is 2.3219

Ans 2.3219

Thnaks Sona22 fr your prompt feedback!

Yes, 2.3219 is indeed a solution to the equation. BUT there are others! The question was saying all the possible solutions for x (not necessarily real numbers!). may be you can retreive them?