Question

Hello, I need some help with this precalculus question for my homework, please HW Q7

Answer 1

Given the equation:

`log_5(x+34)+log_5(x+134)=5`

Apply the product property of logarithm.

Since they have the same bases, we have:

`log_5((x+34)(x+134))=5`

Now, expand using FOIL method and apply distributive property:

`\begin{gathered} log_5(x(x+134)+34(x+134))=5 \\ \\ log_5(x(x)+x(134)+34(x)+134(34))=5 \\ \\ log_5(x^2+134x+34x+4556)=5 \\ \\ log_5(x^2+168x+4556)=5 \end{gathered}`

Rewrite the equation using the definition of logarithm:

`\begin{gathered} x^2+168x+4556=5^5 \\ \\ x^2+168x+4556=3125 \end{gathered}`

Equate to zero.

Subtract 3125 from both sides:

`\begin{gathered} x^2+168x+4556-3125=3125-3125 \\ \\ x^2+168x+1431=0 \end{gathered}`

Factor the left side of the equation using AC method.

Find a pair of numbers whose sum is 168 and product is 1431

We have:

9 and 159

We nor have the factored form:

`(x+9)(x+159)=0`

Equate each factor to zero and solve for x:

`\begin{gathered} x+9=0 \\ Subtract\text{ 9 from both sides:} \\ x+9-9=0-9 \\ x=-9 \\ \\ \\ x+159=0 \\ \text{ Subtract 159 from both sides:} \\ x=-159 \end{gathered}`

Hence, we have the solutions:

x = -9 and -159

To find the real solution substitute each value of x in the equation to determine which solution makes the equation true.

When x = -9:

`\begin{gathered} log_(5)(x+34)+log_(5)(x+134)=5 \\ \\ log_5(-9+34)+log_5(-9+134)=5 \\ \\ log_5(25)+log_5(125)=5 \\ \\ log_5(5^2)+log_5(5^3)=5 \\ \\ 2+3=5 \\ \\ 5=5 \end{gathered}`

-9 makes the logarithm equation true.

When x = -159:

`\begin{gathered} log_(5)(x+34)+log_(5)(x+134)=5 \\ \\ log_5(-159+34)+log_5(-159+134)=5 \\ \\ log_5(-125)+log_5(-25)=5 \end{gathered}`

Since we have the log of negative numbers, the equation is undefined when x = -159.

Therefore, the correct solution is x = -9.

The solution set is {-9}

ANSWER:

A. The solution set is {-9}