How do you solve ${log}_{2} x + {log}_{2} (x - 2) = 3$ $log_2x + log_2(x-2) = 3$ ?

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How do you solve ${log}_{2} x + {log}_{2} (x - 2) = 3$ $log_2x + log_2(x-2) = 3$ ?

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Answer 1 · 2016-04-21T18:35:39

$x = 4$ $x=4$

Explanation:

Use Property: ${log}_{b} (x y) = {log}_{b} x + {log}_{b} y and {log}_{b} y = x \Leftrightarrow x^{b} = y$ $log_b(xy)=log_bx+log_by and log_by=x iff x^b=y$

${log}_{2} (x (x - 2)) = 3$ $log_2(x(x-2))=3$

$2^{3} = x^{2} - 2 x$ $2^3=x^2-2x$

$8 = x^{2} - 2 x$ $8=x^2-2x$

$0 = x^{2} - 2 x - 8$ $0=x^2-2x-8$

$0 = (x - 4) (x + 2)$ $0=(x-4)(x+2)$

$x - 4 = 0 or x + 2 = 0$ $x-4=0 or x+2=0$

$x = 4 or x = - 2$ $x=4 or x=-2$

$x = 4$ $x=4$

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How do you solve ${log}_{2} x + {log}_{2} (x - 2) = 3$ $log_2x + log_2(x-2) = 3$ ?

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Explanation:

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How do you solve ${log}_{2} x + {log}_{2} (x - 2) = 3$ $log_2x + log_2(x-2) = 3$ ?