What did I do wrong (Logarithmic Equation)?

Question

Given $\log_5 (x+35) + \log_5(x+15)=3,$ I did the following:

$\log_5 (x+35) + \log_5(x+15)=3$

$\log_5(x^2+50x+525)=3$

$5^{\log_5(x^2+50x+525)}=5^3$

$x^2+50x+525=125$

$x^2+50x+400=0$

$(x+10)(x+40)=0$

$x=-10, x=-40$

$Domain: (x+35)>0 , (x+15)>0$

$x=-10$

Now this is correct. However, I first attempted to solve it another way which got me an incorrect answer. I knew ahead of time by using the first method above I would have to multiply binomials and then factor which I did not want to do (yes I realize how easy the polynomial turned out to be but I did not know at the time) so I intended to move a $\log$ to the other side and cancel using $x^{\log_x(a)}=a$ Clearly I am doing something wrong/misunderstanding certain rules in the following method:

$\log_5 (x+35) + \log_5(x+15)=3$

$\log_5(x+35) =3-\log_5(x+15)$

$5^{\log_5(x+35)}=5^3-5^{\log_5(x+15)}$

$x+35=125-(x+15)$

$x+35=125-x-15$

$2x=75$

$x=75/2$

Clearly 75/2 is not 10,so what have I done wrong? I have been trying to figure it out for some time, Thanks!

Answer 1

You made a mistake after this line.. $$\log_5(x+35) =3-\log_5(x+15)$$ $$(x+35)=5^3\times 5^{-\log_5(x+15)}$$ $$(x+35)=5^3\times (x+15)^{-1}= \frac {5^3 }{(x+15)}$$

You summed instead..

Answer 2

$\log_5(x+35)+\log_5(x+15)=3$

$\log_5(x+35)=3−\log_5(x+15)$

$5^{\log_5(x+35)}=5^{3−\log_5(x+15)} !!!\ne!!! 5^3 - 5^{\log_5(x+15)}$

$5^{\log_5(x+35)}=5^{3−\log_5(x+15)} = \frac {5^3}{5^{\log_5(x+15)}}$

$x + 35 = \frac {125}{x+15}$

$(x+35)(x + 15) = 125$ .... and ... that's not significantly different than the first way...