solve for x : log base 4 (x2 + 3x) - log base 4 (x+5) = 1?

(x^2 + 3x)/(x + 5) = 4, so

x^2 + 3x = 4x + 20, so

x^2 - x - 20 = 0, so

(x - 5)(x + 4) = 0, so

x = 5, or x = -4.

log5 (x - 3) + log5 (x + a million) = log5 (x + 3) (combine logs) log5 ((x - 3)(x + a million)) = log5 (x + 3) (drop the logs) (x - 3)(x + a million) = x + 3 (FOIL) x^2 - 2x - 3 = x + 3 (subtract x from each edge) x^2 - 3x - 3 = 3 (upload 3 to each edge) x^2 - 3x = 6 (take 0.5 of -3, it really is -3/2, and sq. it to 9/4; upload 9/4 to each edge) x^2 - 3x + 9/4 = 6 + 9/4 x^2 - 3x + 9/4 = 24/4 + 9/4 x^2 - 3x + 9/4 = 33/4 (component left edge) (x - 3/2)^2 = 33/4 (sq. root each edge) x - 3/2 = ±?(33)/2 (upload 3/2 to each edge) x = 3/2 ± ?(33)/2 the a danger ideas for x are 3/2 + ?(33)/2 or 3/2 - ?(33)/2. besides the indisputable fact that, once you change 3/2 - ?(33)/2 for x lower back into the equation, you eventually end up with taking the log of a detrimental, which isn't a danger, so 3/2 - ?(33)/2 isn't a answer. therefore 3/2 + ?(33)/2 is your in easy terms answer. answer: x = 3/2 + ?(33)/2 undergo in thoughts that once logs of an identical base are being extra at the same time, you could combine them into one log through multiplying them. as an social gathering, log 6 + log 8 = log 40 8. contained in the equation of your situation, you need to then drop the logs on each edge because the equation became basically log = log and both logs were of an identical base.

4 ^ (log4 (x^2 + 3x) - log4 (x + 5)) = 4^1

(x^2 + 3 x) / (x + 5) = 4

x^2 + 3x = 4x + 20

x^2 - x - 20 = 0

(x - 5)(x + 4) = 0

x = 5 or x = -4

log_4 (x² + 3x) - log_4 (x+5) = 1

log_4 [(x² + 3x) / (x+5)] = 1

(x² + 3x) / (x+5) = 4^1

x² + 3x = 4(x+5)

x² - x - 20 = 0

x = 5 . . . or . . . x = - 4

Alejandra