Question

Solve x for log(x+5) + log(x-16)=2

Answer 1

Answer:

x = 20

Step-by-step explanation:

Using the rule of logarithms

log x + log y = log xy

$log_{b}$ x = n ⇔ x = $b^{n}$

Given

log(x + 5) + log(x - 16) = 2

log (x + 5)(x - 16) = 2, hence

(x + 5)(x - 16) = 10² = 100 ← expand left side

x² - 11x - 80 = 100 ( subtract 100 from both sides )

x² - 11x - 180 = 0 ← in standard form

(x - 20)(x + 9) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 20 = 0 ⇒ x = 20

x + 9 = 0 ⇒ x = - 9

However, x > 0 ⇒ x = 20

Answer 2

Answer:  x = 20

Step-by-step explanation:

log(x + 5) + log(x - 16) = 2

Use the log rule for condensing: addition becomes multiplication

log[(x + 5)(x - 16)] = 2

Use the rule for eliminating log. Reminder that this is log base 10.

(x + 5)(x - 16) = 10²

Simplify

x² - 11x - 80 = 100

x² - 11x - 180 = 0

Factor the quadratic equation and solve for x

(x - 20)(x + 9) = 0

x = 20  x = -9

Check to see if solutions are valid (both logs must be greater than 0)

         x + 5 > 0     and        x - 16 > 0

20:   20 + 5 > 0    and       20 - 16 > 0          Both are TRUE

-9:     -9 + 5 > 0    FALSE! not a solution!

So, x = 20 is the only valid solution.