How do you solve for x in ${log}_{2} (x - 3) = 2 - {log}_{2} (x - 6)$ $log_2(x-3)=2-log_2(x-6)$ ?

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How do you solve for x in ${log}_{2} (x - 3) = 2 - {log}_{2} (x - 6)$ $log_2(x-3)=2-log_2(x-6)$ ?

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Answer 1 · 2018-04-08T01:42:28

$x = 2, 7$ $x=2,7$

Explanation:

Given ${log}_{2} (x - 3) = 2 - {log}_{2} (x - 6)$ $log_2 (x-3)=2-log_2 (x-6)$
$or, {log}_{2} (x - 3) + log (x - 6) = 2$ $or, log_2 (x-3)+log (x-6)=2$
$or, {log}_{2} [(x - 3) (x - 6)] = 2$ $or, log_2 [(x-3)(x-6)]=2$
$or, {log}_{2} (x^{2} - 9 x + 18) = 2$ $or, log_2 (x^2-9x+18)=2$
$or, {log}_{2} (x^{2} - 9 x + 18) = {log}_{2} (4)$ $or, log_2 (x^2-9x+18)=log_2 (4)$
By one to one property,
$(x^{2} - 9 x + 18) = 4$ $(x^2-9x+18)=4$
$or, x^{2} - 9 x + 18 - 4 = 0$ $or, x^2-9x+18-4=0$
$or, x^{2} - 9 x + 14 = 0$ $or, x^2-9x+14=0$
$or, x^{2} - 7 x - 2 x + 14 = 0$ $or, x^2-7x-2x+14=0$
$or, (x - 2) (x - 7) = 0$ $or, (x-2)(x-7)=0$
$x = 2, 7$ $x=2,7$
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How do you solve for x in ${log}_{2} (x - 3) = 2 - {log}_{2} (x - 6)$ $log_2(x-3)=2-log_2(x-6)$ ?

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