How do you solve #ln(2x^2-2)-ln9=ln80#?

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Answer 1 · 2016-08-18T15:00:45

The Soln.# : x=+-19#.

#ln(2x^2-2)-ln9=ln80#

#:. ln(2x^2-2)=ln9+ln80=ln(9*80)#

Since, #ln# fun. is #1-1#, we have,

#2x^2-2=9*80#

#rArr 2(x^2-1)=9*80#

#rArr x^2-1=9*40=360#

#rArr x^2=361#

#rArr x=+-sqrt361=+-19#

#x=+19, and, x=-19# satisfy the given eqn.

Hence, the Soln. #x=+-19#.

Answer 2 · 2016-08-18T17:37:11

#x = +-19#

If the logs are being subtracted, the numbers are being divided.

We can condense two ln terms into one.

#ln(2x^2-2) - ln9 =ln80#

#ln((2x^2-2)/9) = ln 80#

#:. (2x^2-2)/9 = 80" if "ln(a/b) = ln c rArr a/b = c#

#2x^2-2 = 720#

#2x^2= 722#

#x^2 = 361#

#x= +-sqrt361#

#x = +-19#