How do you solve ${log}_{4} 7 + 2 {log}_{4} x = {log}_{4} 2$ $log_4 7 + 2 log_4 x = log_4 2$ ?

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How do you solve ${log}_{4} 7 + 2 {log}_{4} x = {log}_{4} 2$ $log_4 7 + 2 log_4 x = log_4 2$ ?

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Answer 1 · 2015-12-14T11:34:13

$x = \sqrt{\frac{2}{7}}$ $x=sqrt(2/7)$

Explanation:

Given:
$XXX {log}_{4} (7) + 2 {log}_{4} (x) = {log}_{4} (2)$ $color(white)("XXX")log_4(7)+2log_4(x)=log_4(2)$

Remember
$XXX$ $color(white)("XXX")$ log multiplication rule: ${log}_{b} (p \cdot q) = {log}_{b} (p) + {log}_{b} (q)$ $log_b(p*q) =log_b(p)+log_b(q)$
$XXX$ $color(white)("XXX")$ log power rule: ${log}_{b} (s^{t}) = t \cdot {log}_{b} (s)$ $log_b(s^t) = t*log_b(s)$

Therefore the given equation can be rewritten as
$XXX {log}_{4} (7 x^{2}) = {log}_{4} (2)$ $color(white)("XXX")log_4(7x^2)=log_4(2)$

from which it follows that
$XXX 7 x^{2} = 2$ $color(white)("XXX")7x^2=2$

$XXX x^{2} = \frac{2}{7}$ $color(white)("XXX")x^2=2/7$

$XXX x = \sqrt{\frac{2}{7}}$ $color(white)("XXX")x=sqrt(2/7)$
$XXXXXX$ $color(white)("XXXXXX")$ we can ignore the negative root as extraneous since $\sqrt{x}$ $sqrt(x)$ requires $x \geq 0$ $x>=0$ for Real solutions

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How do you solve ${log}_{4} 7 + 2 {log}_{4} x = {log}_{4} 2$ $log_4 7 + 2 log_4 x = log_4 2$ ?

1 Answer

Explanation:

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How do you solve ${log}_{4} 7 + 2 {log}_{4} x = {log}_{4} 2$ $log_4 7 + 2 log_4 x = log_4 2$ ?