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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!

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  • $\begingroup$ 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). $\endgroup$ – Dominated Convergence Theorem Apr 25 '18 at 21:14
  • $\begingroup$ Your third line: If $x = a + b$ then $5^x \ne 5^a + 5^b$. $5^x = 5^{a + b} = 5^a5^b$. So $5^{\log_5(x+35)} = \frac {5^3}{5^{\log_5(x+15)}}$ $\endgroup$ – fleablood Apr 25 '18 at 21:18
  • $\begingroup$ $5^{a- b}=\frac{5^a}{5^b}$ NOT $5^a- 5^b$. $\endgroup$ – user247327 Apr 25 '18 at 21:19
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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..

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

$\log_5(x+35)=3−\log_5(x+15)$

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

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

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

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

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  • $\begingroup$ 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. $\endgroup$ – Sphygmomanometer Apr 25 '18 at 21:29

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