How do you solve ${log}_{10} (x^{2} + 1) = 1$ $log_10(x^2+1)=1$ ?

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Answer 1 · 2016-11-19T09:22:49

$x = \pm 3$ $x=+-3$

${log}_{a} b = c \Rightarrow a^{c} = b$ $log_ab=c=>a^c=b$

${log}_{10} (x^{2} + 1) = 1 \Rightarrow 10^{1} = x^{2} + 1$ $log_10(x^2+1)=1=>10^1=x^2+1$

$x^{2} + 1 = 10$ $x^2+1=10$

$x^{2} = 9$ $x^2=9$
$x = \sqrt{9} = \pm 3$ $x=sqrt9=+-3$

How do you solve log10(x2+1)=1log_10(x^2+1)=1?