anonymous
  • anonymous
Evalute log 2(x-2) + log 2 (x+5)=3
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Zarkon
Nnesha
  • Nnesha
familiar with the log properties ?
anonymous
  • anonymous
yes

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Nnesha
  • Nnesha
okay so apply one of them at left side \[\huge\rm \log_2(x-2) \color{reD}{+} \log_2 (x+5)=3\] there is plus sign whichnoe one would you use ? quotient or product property ?
anonymous
  • anonymous
product
Nnesha
  • Nnesha
yes \[\huge\rm log_b x + \log_b y = \log_b(x \times y)\] try to apply this on ur original question pretty sure you can do it!
anonymous
  • anonymous
x^2+3x-10
Nnesha
  • Nnesha
yes \[\huge\rm log_2 (x^2+3x-10)=3\] now we have to convert log to exponential form
anonymous
  • anonymous
x^2+3x=13
Nnesha
  • Nnesha
how did you get 13 ?
Nnesha
  • Nnesha
ohh well log_2 is still there \[\huge\rm log_b x + \log_b y = \color{ReD}{\log_b}(x \times y)\] \[\huge\rm\color{reD}{ log_2} (x^2+3x-10)=3\] log_b is a common factor
anonymous
  • anonymous
now what do i do?
UnkleRhaukus
  • UnkleRhaukus
now, exponentiate both sides of the equation, with the base 2
anonymous
  • anonymous
What does that mean?
Nnesha
  • Nnesha
|dw:1439613747571:dw| convert log to exponential form
UnkleRhaukus
  • UnkleRhaukus
from\[LHS = RHS\] \[2^{LHS} = 2^{RHS}\]
anonymous
  • anonymous
ok hold on let me try..
anonymous
  • anonymous
y=2^x ?? Im so confused.
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\end{align} \]
Nnesha
  • Nnesha
that was an example |dw:1439613949291:dw|
anonymous
  • anonymous
Yay! :) Would that be the final answer?
UnkleRhaukus
  • UnkleRhaukus
solve the quadratic
anonymous
  • anonymous
I can do that :) Please hold on
anonymous
  • anonymous
(x-5)(x+2)
anonymous
  • anonymous
wait! I factored wrong! Ughhh hold again
anonymous
  • anonymous
(x+5)(x-2)
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\\ x^2+3x-18&=0\\ &= \end{align}\]
anonymous
  • anonymous
(x+6)(x-3)
UnkleRhaukus
  • UnkleRhaukus
\[x^2+3x-18=0\\ (x+6)(x-3)=0\] good, so what values of \(x\) do you get?
anonymous
  • anonymous
x=-6,3
UnkleRhaukus
  • UnkleRhaukus
If we try plugging the first of these (\(x=-6\)) back into the original equation \[\log _2(x-2) +\log_2 (x+5)=3\] we get\[\log _2(-6-2) +\log_2 (-6+5)=3\\\log _2(-8) +\log_2 (-1)=3\] .... logs of negative numbers... hmm doesn't make much sense. If we plugging the second of these (\(x=3\)) back into the original equation \[\log _2(x-2) +\log_2 (x+5)=3\] we get\[\log _2(3-2) +\log_2 (3+5)=3\\\log _2(1) +\log_2 (8)=3\\0+\log_2 (2^3)=3\\3\log_2 (2)=3 \\3=3\] which is always true. so this is our solution
anonymous
  • anonymous
Thanks :)!! I understand it now!!!
anonymous
  • anonymous
If I post another question, will u answer it?
anonymous
  • anonymous
solve 7-7x=(3x+2)(x-1) and check for extraneous solutions
anonymous
  • anonymous
7 - 7x = 3x² - x - 2 3x² + 6x - 9 = 0 3(x + 3)(x - 1) = 0 x = -3, 1
anonymous
  • anonymous
Would it be extraneous? Thanks geny55!!! :D
anonymous
  • anonymous
medal please?
anonymous
  • anonymous
I dont thnk it's extraneous? am i right?

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