Japanese | Can you solve this ? | Math Olympiad 

Let y=10^x, then the equation becomes y + y^2 = y^3 or y(y^2 - y - 1)=0. So, x=log[(1+sqr(5))/2]

Substitution with t=10^x leads to the cubic equation t+t^2=t^3 or t^3-t^2-t=0 we can factorize the right side to t^3-t^2-t=t*(t^2--t--1)=t*(t-(sqrt(5)-1)/2)*(t(sqrt(5)+1)/2) and we get t*(t-(sqrt(5)1)/2)*(t---(sqt/5)+1)2)=0 with the 3 solutions t=(1--sqrt(5))/2 or t=0 or t=(1+sqrt(5))/2. Since 10^x>0 for every real x, t=0 and 1--sqrt(5))/2 can be rejected. reverting the substitution leads to 10^x=(1+sqrt(5))/2. With taking ogarithon both side, we get x=llog((1+sqrt(5))/2).

Formulaic: ln(golden ratio)/ln(10).

Got it. It's easy

X=O.209

Забавное решение. А, что 100 это уже не 10^2, а 1000 соответственно не 10^3? Получаем кубическое уравнение вида у^3 - у^2- у=0

😮

100^х = 10^2х 1000^х = 10^3х

решается устно, да

biedronka WON

asnwer=10x isit

Too confusing .........

Ez it’s 0