Question

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

Accepted Answer

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.

Answer

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

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