Help solving the following equation? ${log}_{7} (x - 2) + {log}_{7} (x + 4) = 2 {log}_{7} 4$ $log_7(x-2) + log_7(x+4)=2log_7 4$ Please also check for extraneous solutions.

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Help solving the following equation? ${log}_{7} (x - 2) + {log}_{7} (x + 4) = 2 {log}_{7} 4$ $log_7(x-2) + log_7(x+4)=2log_7 4$ Please also check for extraneous solutions.

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Answer 1 · 2016-12-10T13:29:46

${log}_{7} (x - 2) + {log}_{7} (x + 4) = 2 {log}_{7} 4$ $log_7(x-2) + log_7(x+4)=2log_7 4$

$\Rightarrow {log}_{7} [(x - 2) (x + 4)] = {log}_{7} 4^{2}$ $=>log_7[(x-2)(x+4)]=log_7 4^2$

$\Rightarrow [(x - 2) (x + 4)] = 4^{2}$ $=>[(x-2)(x+4)]= 4^2$

$\Rightarrow x^{2} + 2 x - 8 - 16 = 0$ $=>x^2+2x-8-16=0$

$\Rightarrow x^{2} + 2 x - 24 = 0$ $=>x^2+2x-24=0$

$\Rightarrow x^{2} + 6 x - 4 x - 24 = 0$ $=>x^2+6x-4x-24=0$

$\Rightarrow x (x + 6) - 4 (x + 6) = 0$ $=>x(x+6)-4(x+6)=0$

$\Rightarrow (x + 6) (x - 4) = 0$ $=>(x+6)(x-4)=0$

So $x = - 6 and x = 4$ $x= -6 and x= 4$

for $x = - 6 \to$ $x=-6->$ $x - 2 and x + 4$ $x-2 and x+4$ become negative, so their logarithm is not possible. That means $x = - 6$ $x=-6$ is extraneous solutions.

Hence $x = 4$ $x= 4$ is only solution

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Help solving the following equation? ${log}_{7} (x - 2) + {log}_{7} (x + 4) = 2 {log}_{7} 4$ $log_7(x-2) + log_7(x+4)=2log_7 4$ Please also check for extraneous solutions.

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Help solving the following equation? log7(x−2)+log7(x+4)=2log74log_7(x-2) + log_7(x+4)=2log_7 4 Please also check for extraneous solutions.

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Help solving the following equation? ${log}_{7} (x - 2) + {log}_{7} (x + 4) = 2 {log}_{7} 4$ $log_7(x-2) + log_7(x+4)=2log_7 4$ Please also check for extraneous solutions.