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Unformatted text preview: 1) = log ( 3 x + 12 )
(c) ln10  ln ( 7  x ) = ln x
Solution
(a) 2 log 9 [Solution] [Solution] ( x )  log ( 6 x  1) = 0
9 With this equation there are only two logarithms in the equation so it’s easy to get on one either
side of the equal sign. We will also need to deal with the coefficient in front of the first term. log 9 ( x) 2 = log 9 ( 6 x  1) log 9 x = log9 ( 6 x  1)
Now that we’ve got two logarithms with the same base and coefficients of 1 on either side of the
equal sign we can drop the logs and solve. x = 6x 1
1 = 5x Þ x= 1
5 Now, we do need to worry if this solution will produce any negative numbers or zeroes in the
logarithms so the next step is to plug this into the original equation and see if it does. © 2007 Paul Dawkins 302 http://tutorial.math.lamar.edu/terms.aspx College Algebra æ 1ö
æ 1ö
æ æ1ö ö
æ1ö
2 log9 ç
÷  log9 ç 6 ç ÷  1÷ = 2 log9 ç
÷  log9 ç ÷ = 0
ç 5÷
ç 5÷
è5ø
è è5ø ø
è
ø
è
ø
Note that we don’t need to go all the way out with the check here. We just need to make sure that
once we plug in the x we don’t have any negative numbers or zeroes...
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 Spring '12
 MrVinh

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