We could subtract log 5 both
sides:
log 3x = log (x+7) - log
5
We could use the quotient property of the
logarithms:
log 3x = log
[(x+7)/5]
We'll use the one to one
property:
3x = (x+7)/5
We'll
cross multiply:
15x =
x+7
We'll isolate x to the left
side:
14x = 7
We'll divide by
14 both sides:
x = 7/14
x =
1/2
Since the value of x is positive, the solution of the
equation is admissible and it is x = 1/2.
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