How do you solve ${log}_{6} (x + 1) + {log}_{6} (x - 4) = 1$ $log_6 (x+ 1)+ log_6 (x-4)= 1$ ?

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How do you solve ${log}_{6} (x + 1) + {log}_{6} (x - 4) = 1$ $log_6 (x+ 1)+ log_6 (x-4)= 1$ ?

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Answer 1 · 2016-07-30T22:39:11

$x = 5$ $x=5$

Explanation:

Given:

${log}_{6} (x + 1) + {log}_{6} (x - 4) = 1$ $log_6(x+1)+log_6(x-4) = 1$

We require:

$6 = 6^{1} = 6^{{log}_{6} (x + 1) + {log}_{6} (x - 4)} = (x + 1) (x - 4) = x^{2} - 3 x - 4$ $6 = 6^1 = 6^(log_6(x+1)+log_6(x-4)) = (x+1)(x-4) = x^2-3x-4$

Subtract $6$ $6$ from both ends to get:

$0 = x^{2} - 3 x - 10 = (x - 5) (x + 2)$ $0 = x^2-3x-10 = (x-5)(x+2)$

So $x = 5$ $x=5$ or $x = - 2$ $x=-2$

If $x = 5$ $x=5$ then:

${log}_{6} (x + 1) + {log}_{6} (x - 4) = {log}_{6} (6) + {log}_{6} (1) = 1 + 0 = 1$ $log_6(x+1)+log_6(x-4) = log_6(6)+log_6(1) = 1+0 = 1$

So $x = 5$ $x=5$ is a solution of the original equation.

If $x = - 2$ $x=-2$ then (even if we allow Complex logarithms):

${log}_{6} (x + 1) + {log}_{6} (x - 4)$ $log_6(x+1)+log_6(x-4)$

$= {log}_{6} (- 1) + {log}_{6} (- 6)$ $= log_6(-1)+log_6(-6)$

$= {log}_{6} (1) + \frac{π i}{ln (6)} + {log}_{6} (6) + \frac{π i}{ln (6)}$ $= log_6(1)+(pi i)/ln(6) + log_6(6)+(pi i)/ln(6)$

$= 1 + \frac{2 π i}{ln (6)} \neq 1$ $= 1+(2pi i)/ln(6) != 1$

So $x = - 2$ $x=-2$ is not a solution of the original equation.

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How do you solve ${log}_{6} (x + 1) + {log}_{6} (x - 4) = 1$ $log_6 (x+ 1)+ log_6 (x-4)= 1$ ?

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How do you solve log_6 (x+ 1)+ log_6 (x-4)= 1 log6(x+1)+log6(x−4)=1 log_6 (x+ 1)+ log_6 (x-4)= 1 ?

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How do you solve ${log}_{6} (x + 1) + {log}_{6} (x - 4) = 1$ $log_6 (x+ 1)+ log_6 (x-4)= 1$ ?