How do you solve $ln x + ln (x + 2) = 4$ $lnx+ln(x+2)=4$ ?

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Answer 1 · 2015-12-15T21:45:12

$x = - 1 + \sqrt{e^{4} + 1}$ $x=-1+sqrt(e^4+1)$

Use the following logarithm rule: ${log}_{a} b + {log}_{a} c = {log}_{a} (b c)$ $log_a b+log_a c=log_a(bc)$

$ln (x (x + 2)) = 4$ $ln(x(x+2))=4$

$ln (x^{2} + 2 x) = 4$ $ln(x^2+2x)=4$

Recall that $ln x = {log}_{e} x$ $lnx=log_ex$ .

$e^{ln (x^{2} + 2 x)} = e^{4}$ $e^(ln(x^2+2x))=e^4$

$x^{2} + 2 x = e^{4}$ $x^2+2x=e^4$

$x^{2} + 2 x + 1 = e^{4} + 1$ $x^2+2x+1=e^4+1$

${(x + 1)}^{2} = e^{4} + 1$ $(x+1)^2=e^4+1$

$x + 1 = \pm \sqrt{e^{4} + 1}$ $x+1=+-sqrt(e^4+1)$

$x = - 1 \pm \sqrt{e^{4} + 1}$ $x=-1+-sqrt(e^4+1)$

Plug in both values for $x$ $x$ . Notice that only the positive since version works since it's impossible to take the logarithm of a negative number.

$x = - 1 + \sqrt{e^{4} + 1} \approx 6.456$ $x=-1+sqrt(e^4+1)~~6.456$

graph{ln(x)+ln(x+2)-4 [-7.54, 20.94, -6.05, 8.19]}

How do you solve lnx+ln(x+2)=4lnx+ln(x+2)=4?