Given the basis x and x+2 of logarithms find all real solutions of equation 5 - 2[log(x+2) x + logx (x+2)] = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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For the beginning, we'll re-arrange the terms of the equation, by shifting 5 to the right side.

-2[log(x+2) x + logx (x+2)] = -5

We'll divide by -2 both sides:

[log(x+2) x + logx (x+2)] = 5/2

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

x>0

x+2>0

x>-2

x different from 1

x + 2 different from 1

x different from -1.

The interval of admissible values for x is (0 ; +infinite)-{1}.

Now, we'll solve the equation. We'll start by changing the base with by the argument, in the 1st term:

log(x+2) x = 1/logx (x+2)

We'll substitute logx (x+2) by the variable t.

We'll write the equation in t:

1/t + t = 5/2

We'll multiply by 2t both sides:

2 + 2t^2 = 5t

We'll move all terms to one side:

2t^2 - 5t + 2 = 0

We'll solve the equation applying quadratic formula:

t1 = [5+sqrt(25 - 16)]/4

t1 = (5+3)/4

t1 = 2

t2 = (5-3)/4

t2 = 1/2

But logx (x+2) = t1 => logx (x+2) = 2

We'll take antilogarithms:

x + 2 = x^2

We'll subtract x+2 both sides:

x^2 - x - 2 = 0

We'll apply again quadratic formula, to determine x1 and x2:

x1 = [1+sqrt(1+8)]/2

x1 = (1+3)/2

x1 = 2

x2 = (1-3)/2

x2 = -1

We'll reject the second value of x since it doesn't respect the constraints of existence of logarithms.

Now, we'll put

But logx (x+2) = t2 => logx (x+2) = 1/2

We'll take antilogarithms:

x + 2 = sqrtx

We'll raise to square both sides:

(x+2)^2 = x

We'll expand the square:

x^2 + 4x + 4 = x

x^2 + 3x + 4 = 0

We'll apply quadratic formula:

x1 = [-3+sqrt(9-16)]/2

Since sqrt-7 is undefined, the equation x^2 + 3x + 4 = 0 has no real solutions.

The only real solution of the equation is: x = 2.

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