Nice Exponential Equation Problem | Math Olympiad Challenge

[@waltenegus]

if x = b^50, then I got x = 1/2 + 3/2, and x = 1/2 - 3/2, which results in x = -1, x = 2

[@MathswithChinwendu]

👍👍

[@childofGOD1962]

b=0,04 so 100 b = 4 and then 50 b = 2 so 4-2=2

[@MathswithChinwendu]

👍

[@arkadeusz91]

Ummm... That was not the equation to solve... it's not 100b+50b=2, it's b^100+b^50=2 (or b to the 100th power plus b to the 50th power equals 2). Nonetheless you solved the first equation correctly :)

[@stpat7614]

I get three solutions, not two:
b^100 + b^50 = 2
b^100 + b^50 - 2 = 2 - 2
b^100 + b^50 - 2 = 0
b^(50*2) + b^(50) - 2 = 0
(b^50)^2 + b^50 - 2 = 0
Let t = b^50
t^2 + t - 2 = 0
t^2 + (-1t + 2t) + (-1)*(2) = 0
(t^2 - 1t) + (2t - 2) = 0
t(t - 1) + 2(t - 1) = 0
(t - 1)(t + 2) = 0
t - 1 = 0, or t + 2 = 0
t - 1 + 1 = 0 + 1, or t + 2 - 2 = 0 - 2
t = 1, or t = -2
b^50 = 1, or b^50 = -2
Suppose b^50 = 1
(b^50)^(1/2) = +/- 1^(1/2)
b^(50/2) = +/- 1^(1/2)
b^25 = +/- 1
b^25 = 1, or b^25 = -1
(b^25)^(1/25) = 1^(1/25), or (b^25)^(1/25) = (-1)^(1/25)
b^(25/25) = 1, or b^(25/25) = (-1)
b = 1, or b = -1
Suppose b^50 = -2
log(b^50) = log(-2)
50*log(b) = log(-2)
50*log(b)/50 = log(-2)/50
log(b) = log(-2)/50
log(b) = log(2*[-1])/50
log(b) = log(2*i^2)/50
log(b) = (log[2] + log[i^2]) / 50
log(b) = (log[2] + 2*log[i]) / 50
e^log(b) = e^([log(2) + 2*log(i)] / 50)
b = e^([log(2) + 2*log(i)] / 50)
b1 = 1
b2 = -1
b3 = e^([log(2) + 2*log(i)] / 50)

[@MathswithChinwendu]

Wow thank you so much for sharing this

[@stpat7614]

@@MathswithChinwendu Was I right? Isn't -1 also a solution?

[@stpat7614]

Can't b also be equal to -1?

[@MathswithChinwendu]

It can because it satisfies the equation

[@stpat7614]

@@MathswithChinwendu I can't seem to find the error in your formula.