What is the solution set of ${log}_{7} (x + 3) + {log}_{7} (x - 5) = 1$ $log_7 (x+3) + log_ 7 (x-5) = 1$ ?

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Answer 1 · 2018-03-22T03:33:31

$x = 1 + \sqrt{23} \approx 5.7958$ $x = 1+sqrt23 approx 5.7958$

${log}_{7} (x + 3) + {log}_{7} (x - 5) = 1$ $log_7 (x+3) + log_7 (x-5) =1$

Applying ${log}_{n} a + {log}_{n} b = {log}_{n} a b$ $log_n a + log_n b = log_n ab$

${log}_{7} (x + 3) (x - 5) = 1$ $log_7 (x+3)(x-5) = 1$

Now, $1 = {log}_{7} 7$ $1 = log_7 7$

$:. log_7 (x+3)(x-5) = log_7 7$

If $log_n a = log_n b$ then $a=b$

Thus, $(x+3)(x-5) =7$

$x^2 -2x-15 =7$

$x^2-2x-22 =0$

Apply quadratic formula

$x =(2+-sqrt(4+4xx22))/2$

$= 1+- sqrt92/2 =1+- (2sqrt23)/2$

$= 1+-sqrt23$

Remember that $log_n (x)$ is undefined for $x<0$

Then, since $sqrt23>1$ , we can reject the negative result.

NB: since $1+sqrt23>5$ , we can retain the positive result

$:. x=1+sqrt23 approx 5.7958$

What is the solution set of log_7 (x+3) + log_ 7 (x-5) = 1log7(x+3)+log7(x−5)=1log_7 (x+3) + log_ 7 (x-5) = 1?