How do you solve #log x = log (2x^2 - 2)#?

1 Answer
Jan 21, 2016

#x=(1+sqrt(17))/4#

Explanation:

If
#color(white)("XXX")log(x)=log(2x^2-2)#
then
#color(white)("XXX")x=2x^2-2#

#rArr 2x^2-x-2=0#

Using the quadratic formula for roots: #x=(-b+-sqrt(b^2-4ac))/(2a)#
gives
#color(white)("XXX")x=(1+-sqrt(17))/4#

Since the argument of #log# can not be negative,
we can eliminate #(1-sqrt(17))/4# as extraneous,
leaving only the answer above.