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KCC's Quizzes AQQ253 about a funny equation

And here is a new challenge! It's easier than the previous one but still...

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  • 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  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?

Reply Children
  • So we can get value of t as (-5+i√79) /2 and  (-5-i√79) /2 so value of x will be log( -5+i√79) /2*log2 and  (-5-i√79) /2*log2.

    Not sure if we can calculate further. 

    If we can please help me know more about that. 

  • Use the Euler transformation:   a + b i   =  r*exp(i*t)    with r^2  = a^ + b^2   and t is the angle with a base "a" and a rise "b". 
    So  ln(a+bi)  =  ln( r*exp(it)) = ln(r) + ln(exp( i*t ))  =  ln(r)  +  i*t  

    (Euler transform is like to transform a point to polar coordinate  from its cartesian coordinates (real and imaginary axis))

  • Yes! @Sina22! You get it. The expressions fo the 2 complex values can be simplified as -2.5 +/- i 4.4444