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

1 Answer
Feb 17, 2016

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.