How do you solve ${log}_{7} 7 (x + 1) + {log}_{7} (x - 5) = 1$ $log_7 7(x+1) + log_7 (x-5)=1$ ?

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Answer 1 · 2015-11-01T10:03:09

$x = 2 \pm 2 \sqrt{2}$ $x = 2 ± 2sqrt 2$

$log a + log b = log (a b)$ $log a + log b = log (ab)$ , then
${log}_{7} [7 (x + 1) (x - 5)] = 1$ $log_7 [7(x+1)(x-5)] = 1$

${log}_{b} a = c \Rightarrow b^{c} = a$ $log_b a = c \Rightarrow b^c = a$ , then
$7^{1} = 7 (x + 1) (x - 5)$ $7^1 = 7(x+1)(x-5)$

$1 = (x + 1) (x - 5)$ $1 = (x+1)(x-5)$ , continue with Baskara

How do you solve log77(x+1)+log7(x−5)=1log_7 7(x+1) + log_7 (x-5)=1?