Question

log10(x)+log10(x²)+...+ log10(x¹⁹)=380
Sugestão: 1+2+3+...+n=(1+n)n/2

Answer 1

$\displaystyle \sf \\ \underline{\text{Propriedade de} \log} : \\\\ \log _{a} x^b = b\cdot \log_{a}x \\\\ temos :\\\\ \log_{10}x+\log_{10}x^2+\log_{10}x^3+...+\log_{10}x^{19} = 380 \\\\ \log_{10}x + 2\cdot \log_{10}x+3\cdot \log_{10}x + ... + 19\cdot \log_{10}x = 380 \\\\ \boxed{\begin{array}{I} \displaystyle \sf \text{Soma dos termos de uma P.A} : \\\\ \displaystyle \sf a_1+a_2+...+a_n = \frac{(a_1+a_n)\cdot n }{2} \end{array} }$

$\displaystyle \sf temos : \\\\ a_ 1 = \log_{10}x \ ; \ a_n = 19\cdot \log_{10}x \ \ ; \ \ n = 19 \\\\ \text{Da{\'i}} :\\\\ \log_{10}x + 2\cdot \log_{10}x+3\cdot \log_{10}x + ... + 19\cdot \log_{10}x = 380 \\\\\\ \frac{\left(\log_{10}x+19\cdot \log_{10}x \right)\cdot 19}{2} = 380 \\\\\\ \frac{20\cdot \log_{10}x\cdot 19}{2} = 20\cdot 19 \\\\\\ \log_{10}x = 2 \\\\ x= 10^2 \\\\ \huge\boxed{\ \sf x = 100\ }\checkmark$