anonymous
  • anonymous
Solve log 2x2 + 2 = 6. Round to the nearest thousandth if necessary.
Mathematics
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anonymous
  • anonymous
Solve log 2x2 + 2 = 6. Round to the nearest thousandth if necessary.
Mathematics
katieb
  • katieb
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Nnesha
  • Nnesha
is it \[\huge\rm log 2x^2 +2 =6\] ?
anonymous
  • anonymous
I'm getting \[50\sqrt{2} \] or just about 70.711
Nnesha
  • Nnesha
is it \[\huge\rm log (2x^2 +2 )=6\] ? or 2 is base ?

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trwatkins1
  • trwatkins1
can help :( ???
trwatkins1
  • trwatkins1
*cant
anonymous
  • anonymous
its um log 2x^2 + 2 = 6 (log and 2x^2 are seperate)
Nnesha
  • Nnesha
oh so this is right \[\huge\rm log (2x^2 +2 )=6\] 2 isn't base
Nnesha
  • Nnesha
alright can you please explain how you got the answer :=) what was ur first step ? :=)
Nnesha
  • Nnesha
hmm \[\huge\rm log_2 (x^2+2)=6\] \[\huge\rm log (2x^2+2)=6\] which is ur question ?
anonymous
  • anonymous
it's the second one
anonymous
  • anonymous
I subtract 2 from both sides log10 (2x^2) + 2 -2 = 6-2 log10 (2x^2)=4 I then used a = logb (b^a) so... 4= log10 (10^4) log10 (2x^2) = log10 (10^4) then all this other stuff/ not sure I want to type it all out lol
anonymous
  • anonymous
triciaal
  • triciaal
|dw:1440784656572:dw|
triciaal
  • triciaal
do you have answer choices?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
let me type them up
triciaal
  • triciaal
still not sure of the original
anonymous
  • anonymous
a. 707.106 b. 1.414 c. 502√ , or about 70.711 d. 52√ , or about 7.071
anonymous
  • anonymous
anonymous
  • anonymous
@welshfella hey can you help?
Nnesha
  • Nnesha
wait is there a parentheses ??
Nnesha
  • Nnesha
\[\huge\rm log (2x^2 +2 )=6\] if ur question is like this then you can't move the 2 from the parentheses it's (2x^2+2)
Nnesha
  • Nnesha
\[\huge\rm log ~2x^2 +2 =6\] if there isn't any parentheses then you can subtract 2
Nnesha
  • Nnesha
so which one is right first one or 2nd one ? is it \[\huge\rm log ~2x^2 +2 =6\] `OR` \[\huge\rm log (2x^2 +2 )=6\]
Nnesha
  • Nnesha
if `2nd` is right then 1st step) convert log to exponential form :=)

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