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juin [17]
3 years ago
12

Log2(x-3)+log2x-log2(x+2)=2

Mathematics
1 answer:
sertanlavr [38]3 years ago
5 0

<u>We are given the equation:</u>

log₂(x-3) + log₂x - log₂(x+2) = 2

<u>When is it defined:</u>

in this equation, log₂(x-3) and log₂(x+2) can only be defined when

x-3 >0 and x+2 > 0

solving for the values of x, we get:

x > 3    and   x > -2

which basically means x > 3

<em>Because we are looking for an inequality which is true for both x>-2 and x>3</em>

Hence, x will have a value greater than 3

<u>Solving for x:</u>

using the product rule <em>[logₐb + logₐc = logₐ(bc)]</em>

log₂[(x-3)(x)] - log₂(x+2) = 2  

using the quotient rule <em>[logₐb - logₐc = logₐ(b/c)]</em>

log₂[(x-3)(x) / (x+2)] = 2

from the property <em>[ aˣ = b  ⇒  logₐb = x]</em>

(x² - 3x) / (x+2) = 2²

x² - 3x = 4x + 8

x² - 7x - 8 = 0

x² + x - 8x - 8 = 0

x(x+1) - 8(x+1) = 0

(x-8)(x+1) = 0

(x-8) = 0    OR    (x+1) = 0

x = 8      OR      x = -1

We know that the equation is defined only for x > 3

We can see that x = 8 satisfies that inequality

Therefore, x = 8

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