What are the real solutions of the equation? log2(x+1)+log2(x-2)=2

We have to find the solution of log(2)(x+1) + log(2)(x-2) = 2 log(2)(x+1) + log(2)(x-2) = 2 use log a + log b = log a*b => log(2) (x+1)(x-2) = 2 => (x+1)(x-2) = 2^2 => x^2 - x - 2 = 4 => x^2 - x - 6 = 0 => x^2 - 3x + 2x - 6 = 0 => x(x - 3) + 2(x - 3) = 0 => (x + 2)(x - 3) = 0 => x = -2 and x = 3 As the log of a negative number is not defined eliminate x = -2. The required solution of the equation is x = 3.

Since the bases of the logarithms are matching, we'll apply the product property: log 2(x+1)+log2(x-2)=log2 [(x+1)(x-2)] We'll re-write the equation: log2 [(x+1)(x-2)] = 2 We'll take antilogarithms: [(x+1)(x-2)] = 2^2 [(x+1)(x-2)] = 4 We'll remove the brakets: x^2 - x - 2 - 4 = 0 x^2 - x - 6 = 0 We'll apply quadratic formula: x1 = [1+sqrt(1 + 24)]/2 x1 = (1 + 5)/2 x1 = 3 x2 = -2 Since the common interval of admissible values for x (values that make the logarithms to exist) is (2 , +infinite), we'll reject the negative value and we'll keep as solution of equation only x = 3.

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What are the real solutions of the equation? log2(x+1)+log2(x-2)=2

Expert Answers

justaguide

| Certified Educator

We have to find the solution of log(2)(x+1) + log(2)(x-2) = 2

log(2)(x+1) + log(2)(x-2) = 2

use log a + log b = log a*b

=> log(2) (x+1)(x-2) = 2

=> (x+1)(x-2) = 2^2

=> x^2 - x - 2 = 4

=> x^2 - x - 6 = 0

=> x^2 - 3x + 2x - 6 = 0

=> x(x - 3) + 2(x - 3) = 0

=> (x + 2)(x - 3) = 0

=> x = -2 and x = 3

As the log of a negative number is not defined eliminate x = -2.

The required solution of the equation is x = 3.

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giorgiana1976 | Student

Since the bases of the logarithms are matching, we'll apply the product property:

log 2(x+1)+log2(x-2)=log2 [(x+1)(x-2)]

We'll re-write the equation:

log2 [(x+1)(x-2)] = 2

We'll take antilogarithms:

[(x+1)(x-2)] = 2^2

[(x+1)(x-2)] = 4

We'll remove the brakets:

x^2 - x - 2 - 4 = 0

x^2 - x - 6 = 0

We'll apply quadratic formula:

x1 = [1+sqrt(1 + 24)]/2

x1 = (1 + 5)/2

x1 = 3

x2 = -2

Since the common interval of admissible values for x (values that make the logarithms to exist) is (2 , +infinite), we'll reject the negative value and we'll keep as solution of equation only x = 3.