Question

A raiz real da equação

       log10(x + 1) + 1 = log10
(x2 + 35) é:

 

Answer 1

$log_{n}n=1$
$log_{x}(a*b)\rightleftharpoons log_{x}a+log_{x}b$
$log_{x}a=log_{x}b\rightleftharpoons a=b$
_____________________

$log_{10}(x+1)+1=log_{10}(x^{2}+35)\\log_{10}(x+1)+log_{10}10=log_{10}(x^{2}+35)\\log_{10}[(x+1)*10]=log_{10}(x^{2}+35)\\log_{10}(10x+10)=log_{10}(x^{2}+35)$

Removendo log dos 2 lados:

$10x+10=x^{2}+35\\0=x^{2}-10x+35-10\\x^{2}-10x+25=0$

Soma das raízes: $S = -b/a=-(-10)/1=10$
Produto das raízes: $P=c/a=25/1$

Raízes: 2 números que quando somados dão 10 e quando multiplicados dão 25

$x' = 5\\x'' = 5$

Resposta: x = 5

Answer 2

$log\ 10(x + 1) + 1 = log\ 10(x^2 + 35)\\\\ log\ 10(x+1)-log\ 10(x^2+35) = -1\\\\ log\ 10\frac{(x+1)}{(x^2+35)} = -1\\\\ 10^-^1 = \frac{(x+1)}{(x^2+35)}\\\\\ \frac{1}{10} = \frac{(x+1)}{(x^2+35)}\\\\\ 10x+10 = x^2+35\\\\\ x^2-10x+25 = 0$

$\Delta = (-10)^2-4(1)(25)\\\\ \Delta = 100-100\\\ \Delta = 0\\\\\ x^i = 5\\\\ x^i^i = 5$

$\boxed{S(5)}$