How do you solve for x?: ${log}_{3} (x - 2) + {log}_{3} (x - 4) = 2$ $log_3 (x-2) + log_3 (x-4) = 2$

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How do you solve for x?: ${log}_{3} (x - 2) + {log}_{3} (x - 4) = 2$ $log_3 (x-2) + log_3 (x-4) = 2$

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Answer 1 · 2015-10-26T00:08:25

I found $x = 6.1623$ $x=6.1623$

Explanation:

We can combine the two logs as:
${log}_{3} [(x - 2) (x - 4)] = 2$ $log_3[(x-2)(x-4)]=2$
Use the definition of log to get:
$(x - 2) (x - 4) = 3^{2}$ $(x-2)(x-4)=3^2$
$x^{2} - 4 x - 2 x + 8 = 9$ $x^2-4x-2x+8=9$
$x^{2} - 6 x - 1 = 0$ $x^2-6x-1=0$
$x_{1, 2} = \frac{6 \pm \sqrt{36 + 4}}{2} =$ $x_(1,2)=(6+-sqrt(36+4))/2=$ two solutions:
$x_{1} = 6.1623$ $x_1=6.1623$
$x_{2} = - 0.1623$ $x_2=-0.1623$ NO

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How do you solve for x?: ${log}_{3} (x - 2) + {log}_{3} (x - 4) = 2$ $log_3 (x-2) + log_3 (x-4) = 2$

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How do you solve for x?: ${log}_{3} (x - 2) + {log}_{3} (x - 4) = 2$ $log_3 (x-2) + log_3 (x-4) = 2$