Given
Now
=
=
=
log (2 + 3) = log 5
log (2 x 3) = log 2 + log 3
log (2 + 3)
log (2 x 3)
By trial and error method, when we substitute
x = 0
Both LHS and RHS are equal.
But at x=-1/3, log x is not defined.
The only admissible value of x is 1.
Taking Log to both sides
we get
n = 8
Given equation 5x-17 = -x+7
Add 1x to each side of the equation
5x-17+x = -x+7+x
6x=24
x=4
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