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

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Answer 1 · 2015-12-15T21:45:12

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

Use the following logarithm rule: $log_a b+log_a c=log_a(bc)$

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

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

Recall that $lnx=log_ex$ .

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

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

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

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

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

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

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

$x=-1+sqrt(e^4+1)~~6.456$

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?