Mathematical Reflections 2017, Issue 3 - Problem J409

Problem:Solve the equation $$\log(1-2^x+5^x-20^x+50^x)=2x.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
The given equation is equivalent to $$1-2^x+5^x-20^x+50^x=10^{2x},$$ i.e. $$1-2^x+10^x-20^x+5^x-10^x+50^x-100^x=0,$$ which gives
$$(1-2^x)(1+5^x)(1+10^x)=0.$$ It follows that $1-2^x=0$, so $x=0$.