Question

How do you solve `log_ 5 (x+4) + log _ 5 (x+1)= 2`?

Answer 1

You can start by simplifying using the log property `log_an + log_am = log_a(n xx m)`

Explanation:

`log_5(x + 4) + log_5(x + 1) = 2`

`log_5(x + 4)(x + 1) = 2`

`log_5(x^2 + 4x + x + 4) = 2`

Convert to exponential form:

`x^2 + 5x + 4 = 25`

`x^2 + 4x - 21 = 0`

We can solve this equation by factoring. This is a trinomial of the form `y = ax^2 + bx + c`. To factor this, you must find two numbers that multiply to c and that add to b. Two numbers that multiply to -21 and that add to 4 are +7 and -3.

`(x + 7)(x - 3) = 0`

`x = -7 and 3`

Since having a negative log is undefined, the only solution to the equation is x = 3.

Hopefully this helps.