Question

log(5)x+log(5)(x+3)=1
log(5)x^2+3x=1
5^1=x^2+3x
x^2+3x-5=0
(x+?)(x-?)=0

Answer 1

Well using the quadratic formula,you could factor

\[x^2+3x-5=0\]

to get,
\[ \left\{\left\{x=\frac{1}{2} \left(-3-\sqrt{29}\right)\right\},\left\{x=\frac{1}{2} \left(-3+\sqrt{29}\right)\right\}\right\} \]

Answer 2

I dont think thats correct...?

Answer 3

log(5)x + log(5)(x+3) = 1
log(5) (x^2 + 3x) = 1
x^2 + 3x = 5
x^2 + 3x - 5 = 0
x = (-3 +/- sqrt(9 + 20)) / 2

foolformath is right..