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Mathematics 12 Online
OpenStudy (anonymous):

solve log 2x + log 12 =3 can someone please explain

OpenStudy (anonymous):

write log 2x + log 12 as one log log 2x*12=3 then apply base 10 exponential to both sides, which cancels the log on the left 2x*12=10^3 should be easy now

OpenStudy (anonymous):

Do you know your laws of logarithms?

OpenStudy (anonymous):

Well the left side can be combined into one log using the law that says \[\log_{a}x+\log_{a}y=\log_{a}xy\]

OpenStudy (anonymous):

Right, so that gives us log 2x*12=3 when there is no base written it's the common log with base 10, to cancel a log out you apply that base exponential to both sides, in this case 10 2x*12=10^3

OpenStudy (anonymous):

so 10^3 to both sides

OpenStudy (anonymous):

No, base 10 exp. to both sides. Before i canceled the log out it would look like this \[10^{\log 2x*12}=10^3\]

OpenStudy (anonymous):

right, so the log and the 10 cancel, leaving you with 2x*12=10^3

OpenStudy (anonymous):

Can you solve now?

OpenStudy (anonymous):

You cant solve 24x=1000?

OpenStudy (anonymous):

what happens to the x thats what confuses me

OpenStudy (anonymous):

multiply 2*12 to get 24 and 10*10*10 to get 1000 then multiply the two together to get x which would be 24,000

OpenStudy (anonymous):

x is what you're solving for... 2*12x=24x 10^3=1000 so, 24x=1000 just divide both sides by 24 and x is solved for

OpenStudy (anonymous):

41.67

OpenStudy (anonymous):

Yep

OpenStudy (anonymous):

thank you for helping

OpenStudy (anonymous):

No problem, if you get familiar with these laws of logarithms it makes these problems a lot easier

OpenStudy (anonymous):

ok thanks

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