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anonymous
 one year ago
Evalute log 2(x2) + log 2 (x+5)=3
anonymous
 one year ago
Evalute log 2(x2) + log 2 (x+5)=3

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1familiar with the log properties ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1okay so apply one of them at left side \[\huge\rm \log_2(x2) \color{reD}{+} \log_2 (x+5)=3\] there is plus sign whichnoe one would you use ? quotient or product property ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1yes \[\huge\rm log_b x + \log_b y = \log_b(x \times y)\] try to apply this on ur original question pretty sure you can do it!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1yes \[\huge\rm log_2 (x^2+3x10)=3\] now we have to convert log to exponential form

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1ohh well log_2 is still there \[\huge\rm log_b x + \log_b y = \color{ReD}{\log_b}(x \times y)\] \[\huge\rm\color{reD}{ log_2} (x^2+3x10)=3\] log_b is a common factor

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0now, exponentiate both sides of the equation, with the base 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What does that mean?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439613747571:dw convert log to exponential form

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0from\[LHS = RHS\] \[2^{LHS} = 2^{RHS}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok hold on let me try..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y=2^x ?? Im so confused.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align}\log_2(x^2+3x10)&=3\\ 2^{\log_2(x^2+3x10)}&=2^3\\ (x^2+3x10)&=2^3\end{align} \]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1that was an example dw:1439613949291:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yay! :) Would that be the final answer?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0solve the quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can do that :) Please hold on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait! I factored wrong! Ughhh hold again

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align}\log_2(x^2+3x10)&=3\\ 2^{\log_2(x^2+3x10)}&=2^3\\ (x^2+3x10)&=2^3\\ x^2+3x18&=0\\ &= \end{align}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+3x18=0\\ (x+6)(x3)=0\] good, so what values of \(x\) do you get?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0If we try plugging the first of these (\(x=6\)) back into the original equation \[\log _2(x2) +\log_2 (x+5)=3\] we get\[\log _2(62) +\log_2 (6+5)=3\\\log _2(8) +\log_2 (1)=3\] .... logs of negative numbers... hmm doesn't make much sense. If we plugging the second of these (\(x=3\)) back into the original equation \[\log _2(x2) +\log_2 (x+5)=3\] we get\[\log _2(32) +\log_2 (3+5)=3\\\log _2(1) +\log_2 (8)=3\\0+\log_2 (2^3)=3\\3\log_2 (2)=3 \\3=3\] which is always true. so this is our solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks :)!! I understand it now!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I post another question, will u answer it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve 77x=(3x+2)(x1) and check for extraneous solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.07  7x = 3x²  x  2 3x² + 6x  9 = 0 3(x + 3)(x  1) = 0 x = 3, 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be extraneous? Thanks geny55!!! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont thnk it's extraneous? am i right?
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