How do you solve $2 {log}_{4} x = 1 + {log}_{4} (x + 8)$ $2 log_4 x =1+ log_4 (x + 8)$ ?

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How do you solve $2 {log}_{4} x = 1 + {log}_{4} (x + 8)$ $2 log_4 x =1+ log_4 (x + 8)$ ?

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Answer 1 · 2016-05-07T01:39:54

Solve using the following properties:
• $a log n = {log n}^{a}$ $alogn = logn^a$
• ${log}_{a} (n) - {log}_{a} (m) = {log}_{a} (\frac{n}{m})$ $log_a(n) - log_a(m) = log_a(n/m)$

Explanation:

${log}_{4} (x^{2}) - {log}_{4} (x + 8) = 1$ $log_4(x^2) - log_4(x + 8) = 1$

${log}_{4} (\frac{x^{2}}{x + 8}) = 1$ $log_4((x^2)/(x + 8)) = 1$

$\frac{x^{2}}{x + 8} = 4$ $(x^2)/(x + 8) = 4$

$x^{2} = 4 (x + 8)$ $x^2 = 4(x + 8)$

$x^{2} = 4 x + 32$ $x^2 = 4x + 32$

$x^{2} - 4 x - 32 = 0$ $x^2 - 4x - 32 = 0$

$(x - 8) (x + 4) = 0$ $(x - 8)(x + 4) = 0$

$x = 8 and x = - 4$ $x = 8 and x = -4$

Checking in the original equation, we find that only $x = 8$ $x = 8$ works, so the $x = - 4$ $x = -4$ is extraneous and you therefore will not include it inside the solution set.

Hopefully this helps!

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How do you solve $2 {log}_{4} x = 1 + {log}_{4} (x + 8)$ $2 log_4 x =1+ log_4 (x + 8)$ ?

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How do you solve $2 {log}_{4} x = 1 + {log}_{4} (x + 8)$ $2 log_4 x =1+ log_4 (x + 8)$ ?