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rosedewittbukater
 4 years ago
Algebra 2 help? An explanation would be great too. Thanks.
Solve log 2x + log 12 = 3. Round to the nearest hundredth if necessary. (1 point)
6,000
24,000
166.67
41.67
rosedewittbukater
 4 years ago
Algebra 2 help? An explanation would be great too. Thanks. Solve log 2x + log 12 = 3. Round to the nearest hundredth if necessary. (1 point) 6,000 24,000 166.67 41.67

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cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1first use the following law of logs: log a + log b = log ab

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1so log 2x + log 12 = log ?

rosedewittbukater
 4 years ago
Best ResponseYou've already chosen the best response.0@cwrw238 I tried doing that and got log 24x = 3, and then tried to reexpand it by getting log 24 + log x = 3. Then I subtracted log 24 from both sides and got log x = 1.619. I don't know if what I did was right and if it is what to do from there?

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1log 24x = 3 the definiitoin of log to base 10 is the power that 10 has to be taken to give the required result in other words if log 100 = 2 then 100 = 10^2 so if log 24x = 3 then 24x = 10^3 so 24x = 1000 x = 1000 / 24

rosedewittbukater
 4 years ago
Best ResponseYou've already chosen the best response.0So is it 41.67?

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.1if you remember log to base10 of 100 = 2 then 100 = 10^2 you will have little trouble with these questions 'log' on its own is assumed to be log to the base 10 but you can have log to base any positive number thers also a special logarithm to the base 'e' called a natural logarithm log e sometimes written 'ln'

rosedewittbukater
 4 years ago
Best ResponseYou've already chosen the best response.0Ok thanks. While you're here, do you know how to use log with a base other than 10 on a graphing calculator? Since it has a base of 10 by default, I'm not sure how to put log with a different base. And I just learned about natural logarithms today so thanks!

rosedewittbukater
 4 years ago
Best ResponseYou've already chosen the best response.0@cwrw238 Do you think you could help me with another problem like this? I did what you said but ended up getting several different answers that didn't work when I checked it.
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