Log(x-46)+log(x-61)=2
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i suppose we are working in base 10 for the logarithm
$\displaystyle \log (x - 46) + \log (x - 61) = 2$
$\displaystyle \Rightarrow \log [(x - 46)(x - 61)] = 2$ ..............since $\displaystyle \log_a X + \log_a Y = \log_a XY$
$\displaystyle \Rightarrow (x - 46)(x - 61) = 10^2$ ...............since $\displaystyle \log_a b = c \implies a^c = b$
this is a quadratic, i hope you can take it from here
this is essentially the same as the question you asked here, what did you not understand?
if you have trouble factoring large numbers, you can always use the quadratic formula
to solve $\displaystyle ax^2 + bx + c = 0$ you use the formula $\displaystyle x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
you have $\displaystyle x^2 - 107x + 2706 = 0$, if you want to factor this, you must come up with two numbers that when multiplied gives you 2706 and when added gives you -107.
0. To me, this reply (to someone who has actually just given you exactly the help you require) is very close to being insulting. Try reading the reply again and thinking about what it says.
1. If there's a particular part of a question that you're stuck on, you should state in the initial post what the question is (which you did) and where you're stuck.
Otherwise people waste time including in their reply things that you've already done.
2. Your problem is solving quadratic equations. You're struggling to solve by factorsing. So Jhevon gave you the quadratic formula which can always be used to solve a quadratic equation.
I doubt you would be given these sorts of questions to solve if you hadn't been taught how to solve quadratic equations. Try using the formula. Post where you get stuck (if you get stuck).
3. Given that solving quadratic equations is an obvious weakness, I'd suggest you thoroughly revise it in your textbook or class notes.