How do you solve $2 {log}_{2} x = {log}_{2} (x - 2) + log (x + 4)$ $2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)$ ?

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How do you solve $2 {log}_{2} x = {log}_{2} (x - 2) + log (x + 4)$ $2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)$ ?

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Answer 1 · 2018-04-29T13:49:40

$2 {log}_{2} x = {log}_{2} (x - 2) + {log}_{e} (x + 4)$ $2log_2x=log_2(x−2)+log_e(x+4)$
$2 \frac{{log}_{e} x}{{log}_{e} 2} = \frac{{log}_{e} (x - 2)}{{log}_{e} (2)} + {log}_{e} (x + 4)$ $2log_ex/log_e2=log_e(x-2)/log_e(2)+log_e(x+4)$
$2 \frac{{log}_{e} x}{{log}_{e} 2} = \frac{{log}_{e} (x - 2) + {log}_{e} (x + 4) {log}_{e} 2}{{log}_{e} 2}$ $2log_ex/log_e2=[log_e(x-2)+log_e(x+4)log_e2]/log_e2$
$2 {log}_{e} x = {log}_{e} (x - 2) + {log}_{e} (x + 4) {log}_{e} 2$ $2log_ex=log_e(x-2)+log_e(x+4)log_e2$
${log}_{e} x^{2} - {log}_{e} (x - 2) = {log}_{e} (x + 4) {log}_{e} 2$ $log_ex^2-log_e(x-2)=log_e(x+4)log_e2$
${log}_{e} (\frac{x^{2}}{x - 2}) = {log}_{e} (x + 4) {log}_{e} 2$ $log_e(x^2/(x-2))=log_e(x+4)log_e2$
$ln (\frac{x^{2}}{x - 2}) = ln (x + 4) ln 2$ $ln(x^2/(x-2))=ln(x+4)ln2$
divide by ${log}_{e} 2$ $log_e2$

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How do you solve $2 {log}_{2} x = {log}_{2} (x - 2) + log (x + 4)$ $2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)$ ?

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How do you solve $2 {log}_{2} x = {log}_{2} (x - 2) + log (x + 4)$ $2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)$ ?