How do you solve #lnx=5-ln2x #?

1 Answer
Apr 11, 2016

Answer:

#x=8.614#, rounded to three decimal places.

Explanation:

To solve #lnx=5-ln2x#, for #x#

Rearranging to bring all terms containing #x# to left hand side of the equation we obtain
#lnx+ln2x=5#
Using the property of logarithms that #log (axxb)=loga+log b# we obtain
#ln(x xx 2x)=5#
or #ln2 x^2=5#

Using the definition of log functions we obtain

#e^5=2x^2#, solving for #x#
#x=+-sqrt(e^5/2)#, ignoring the #-ve # sign as logarithms of a negative number is not defined.

#x=sqrt(e^5/2)#, with a calculator

or #x=8.614#, rounded to three decimal places.