Solve: log3(𝑥 − 1) + log3(𝑥 − 6)= 2

Question

College Algebra

Answer 1

Hello, thank you for taking the time to post your question!

You can start by using the properties of logarithms to simplify this one down! So that becomes

log_3(x – 1)(x – 6) = 2

3^2 = (x – 1)(x – 6)

9 = x^2 – 7x + 6

From there then I would put it into standard form and then using the quadratic formula to solve for the exact value! When I do that I get

x^2 – 7x – 3 = 0

x1 = 7.4051 and x2 = -0.4051

However, since x2 = -0.4051 is not larger than 6 that means that only x1 is a valid solution for this system

So the solution to the equation here is x = 7.4051

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

log₃(x-1) + log₃(x-6) = 2

 

log₃[(x-1)(x-6)] = 2

 

(x-1)(x-6) = 3²

 

x² - 7x + 6 = 9

 

x² - 7x - 3 = 0

 

x = [7 ± √61]/2 ≈ 7.4 or -0.41

 

Since logs of negative numbers are undefined, [7 - √61]/2 must be discarded.

 

Solution:  (7 + √61]/2