United States l Nice Olympiad Math Exponential Problem l find possible values of x?

[@zawatsky]

lgi - неочевидная величина. С таким же успехом ответ может быть 1+lb(-1). У нас в детских садах такие задачки называли загадкой про потолочника-лампососа...

[@zawatsky]

x(lbe+iπ+1)=1; ⇒x=1/(lbe+iπ+1).

[@stpat7614]

Third method with tau? e^(i*tau) = 1 -1 * e^(i*tau) = -1 * 1 -e^(i*tau) = -1 (-2)^x = 2 (-1*2)^x = 2 (-1)^x * 2^x = 2 (-e^[i*tau])^x * 2^x = 2 (-1 * e^[i*tau])^x * 2^x = 2 (-1)^x * (e^[i*tau])^x * 2^x = 2 (i^2)^x * (e^[i*tau])^x * 2^x = 2 ln([i^2]^x * [e^(i*tau)]^x * 2^x) = ln(2) ln(i^[2*x] * e^[i*tau*x] * 2^x) = ln(2) ln(i^[2*x]) + ln(e^[i*tau*x]) + ln(2^x) = ln(2) 2x*ln(i) + (i*tau*x)*ln(e) + x*ln(2) = ln(2) 2x*ln(i) + (i*tau*x)*1 + x*ln(2) = ln(2) x*2*ln(i) + x*i*tau + x*ln(2) = ln(2) x*(2*ln[i] + i*tau + ln[2]) = ln(2) x*(2*ln[i] + i*tau + ln[2])/(2*ln[i] + i*tau + ln[2]) = ln(2)/(2*ln[i] + i*tau + ln[2]) x*1 = ln(2)/(2*ln[i] + i*tau + ln[2]) x = ln(2)/(2*ln[i] + i*tau + ln[2])