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]}