QuestionSolve the equation: $\log _{3}(x-5)+\log _{3}(x-11)=3$ .

Studdy Solution

STEP 1

Assumptions1. We are given the equation $\log{3}(x-5)+\log{3}(x-11)=3$ . We are using the logarithm property $\log_b(a) + \log_b(c) = \log_b(ac)$ 3. We are using the logarithm property $b^{\log_b(a)} = a$

STEP 2

First, we can combine the two logarithms on the left side of the equation using the logarithm property $\log_b(a) + \log_b(c) = \log_b(ac)$ .
$\log{}(x-5)+\log{}(x-11)=\log{}((x-5)(x-11))$

STEP 3

So our equation now becomes $\log{3}((x-5)(x-11))=3$

STEP 4

Next, we can convert the logarithmic equation to an exponential equation using the property $b^{\log_b(a)} = a$ . This gives us $(x-)(x-11)=3^3$

STEP 5

implify the right side of the equation $(x-5)(x-11)=27$

STEP 6

Now, expand the left side of the equation $x^2 -16x +55 =27$

STEP 7

Subtract27 from both sides of the equation to set the equation to zero $x^2 -16x +55 -27 =0$

STEP 8

implify the equation $x^2 -16x +28 =0$

STEP 9

Now, factor the quadratic equation $(x-14)(x-2) =$

STEP 10

Finally, solve for x by setting each factor equal to zero $x-14=0 \quad or \quad x-2=0$

STEP 11

olving for x gives $x=14 \quad or \quad x=$ However, if we substitute $x=$ back into the original equation, we get a negative number inside the logarithm, which is undefined. Therefore, the only solution is $x=14$ .

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