By trial and error method, when we substitute
x = 0
Both LHS and RHS are equal.
log (2 + 3) = log 5
log (2 x 3) = log 2 + log 3
log (2 + 3)
log (2 x 3)
Given
Now
=
=
=
= log 1 + log 1 +.....+log 1
= 0.
But at x=-1/3, log x is not defined.
The only admissible value of x is 1.
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