How do you solve $ln \sqrt{x + 1} = 2$ $lnsqrt(x+1) = 2$ ?

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How do you solve $ln \sqrt{x + 1} = 2$ $lnsqrt(x+1) = 2$ ?

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Answer 1 · 2016-03-05T06:46:34

$\Rightarrow x = e^{4} - 1$ $=>x=e^4-1$

Explanation:

$ln \sqrt{x + 1} = 2$ $lnsqrt(x+1)=2$
$\Rightarrow {log}_{e} \sqrt{x + 1} = 2$ $=>log_esqrt(x+1)=2$
$\Rightarrow \sqrt{x + 1} = e^{2}$ $=>sqrt(x+1)=e^2$
$\Rightarrow (x + 1) = e^{4}$ $=>(x+1)=e^4$
$\Rightarrow x = e^{4} - 1$ $=>x=e^4-1$

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