# What did I do wrong (Logarithmic Equation)?

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!

• In the second attempt, when you make both sides have base 5, it is not true that you have subtraction (you get division because of negative exponent). Apr 25 '18 at 21:14
• Your third line: If $x = a + b$ then $5^x \ne 5^a + 5^b$. $5^x = 5^{a + b} = 5^a5^b$. So $5^{\log_5(x+35)} = \frac {5^3}{5^{\log_5(x+15)}}$ Apr 25 '18 at 21:18
• $5^{a- b}=\frac{5^a}{5^b}$ NOT $5^a- 5^b$. Apr 25 '18 at 21:19

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

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

$\log_5(x+35)=3−\log_5(x+15)$

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

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

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

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

• Your third line clarified what you and user247327 meant before since I was confused. I was not making that mistake (I incorrectly summed as pointed out by Isham) but it helps to think of it in that way of exponent rules and gives me a new way of solving such equations. Apr 25 '18 at 21:29