How do you solve ${log x}^{2} + 1 = 5$ $log x^2+1=5$ ?

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How do you solve ${log x}^{2} + 1 = 5$ $log x^2+1=5$ ?

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Answer 1 · 2016-03-21T19:21:07

$x = 100$ $x=100$

Explanation:

${log x}^{2} + 1 = 5$ $log x^2 +1 =5$
${log x}^{2} = 4$ $logx^2=4$ -> subtract 1 from both sides
$2 log x = 4$ $2logx=4$ ->Use property ${log}_{b} x^{n} = n \cdot {log}_{b} x$ $log_bx^n=n*log_b x$
$log x = 2$ $log x=2$ -> divide both sides by 2
$10^{2} = x$ $10^2=x$ ->use property $y = {log}_{b} x \Leftrightarrow b^{y} = x$ $y=log_bx iff b^y=x$
$x = 100$ $x=100$

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How do you solve ${log x}^{2} + 1 = 5$ $log x^2+1=5$ ?

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How do you solve ${log x}^{2} + 1 = 5$ $log x^2+1=5$ ?