What is x if log2x = log5 + log(x - 24/5) ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we'll impose the constraints of existence of logarithms:

x - 24/5>0

x > 24/5

2x>0

x>0

The common interval of admissible values for x is (24/5 ,+inf.).

Now,we'll could solve the equation using the property of quotient:

log 2x = log 5 - log (x - 24/5)

log 2x  = log [5/(x - 24/5)]

Because the bases of logarithms are matching, we'll apply the one to one property:

2x = 5/(x - 24/5)

We'll cross multiply;

2x(x - 24/5) = 5

We'll remove the brackets:

2x^2 - 48x/5 - 5 = 0

We'll multiply by 5:

10x^2 - 48x - 25 = 0

We'll apply the quadratic formula:

x1 = [48+sqrt(2304+1000)]/40

x1 = 2(24+sqrt826)/40 = 2.637 < 24/5

x2 = 2(24-sqrt826)/40 = -0.2< 24/5

Since the values of x1 and x2 do not belong to the interval of admissible values, the equation has no valid solutions

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