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Published byAlejandro Harcourt Modified over 3 years ago

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1) Solve: log 27 x=-2/3 2) Write in logarithmic form: 49 1/2 = 7 3) Graph y = -2(3) -x 1)x= 27 -2/3 x = 2) ½ = log 49 7 3) = -2(1/3) x

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a 3 a 5 = a 3+5 log a 3 + log a 5 = log a (3*5) (a 3 ) 5 = a 3*5 5log a 3 = log a 3 5 a 5 = a 5-3 a 3 log a 5 – log a 3 = log a (5/3) a 0 = 1Log a 1=0 a x is always positive Log a (~) ~ is always positive!! Adding logs, Adding logs, you multiply them A number in front of log becomes the exponent. Minus logs, minus logs, you divide them Log of 1 is zero, and can’t take log of negative.

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Recall that if log x 5 = log x y then 5=y log x 25 = 2 then 25 = x 2 Goal is to condense logs to just ONE LOG on each side…. 13) log 4 5 + log 4 x = log 4 60 Log 4 5x = log 4 60 5x = 60 x = 12 CHECK answer! {12} 24) log 2 (x+4) – log 2 (x-3)= 3 8(x-3) = x+4 8x-24=x + 4 7x = 28 x=4 CHECK! {4} 18) 3 log 8 2 – log 8 4 = log 8 b log 8 2 3 – log 8 4 = log 8 b log 8 (8/4) = log 8 b Log 8 2 = log 8 b 2 = b Problems taken from Glencoe Algebra II workbook 10.3

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11) log 10 27 = 3 log 10 x 12) 3log 7 4=2 log 7 b Log 10 27 = log 10 x 3 27 = x 3 3 = x {3} log 7 4 3 = log 7 b 2 log 7 64 =log 7 b 2 64 = b 2 + 8 = b {8} Problems taken from Glencoe Algebra II workbook 10.3

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16) Log 2 q – log 2 3 = log 2 715) log 5 y-log 5 8=log 5 1 Log 2 (q/3) = log 2 7 q/3 = 7 q = 21 {21} Log 5 y/8 = log 5 1 y/8 = 1 y = 8 {8} Problems taken from Glencoe Algebra II workbook 10.3

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21) log 3 d + log 3 3 = 3 23) log 2 s + 2 log 2 5 =0 Log 3 (d3) = 3 3d = 3 3 3d = 27 d = 9 Log 2 s + log 2 5 2 = 0 Log 2 (s25) = 0 25s = 2 0 25s = 1 s = 1/25 Problems taken from Glencoe Algebra II workbook 10.3

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20) log 10 x + log 10 (3x – 5) = 219) log 4 x+ log 4 (2x – 3) = log 4 2 Log 4 (x(2x -3)) = log 4 (2) 2x 2 – 3x = 2 2x 2 – 3x – 2 = 0 (2x + 1)(x – 2) = 0 2x + 1= 0 x-2=0 x = -1/2 x =2 {2} Log 10 (x(3x-5)) = 2 3x 2 – 5x = 10 2 3x 2 – 5x = 100 3x 2 – 5x – 100 = 0 (3x -20 )(x+5 )=0 x = 20/3 x = -5 {20/3} Problems taken from Glencoe Algebra II workbook 10.3

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Solving Equations with Logs Day 2. Solving equations with only one logarithm in it: If it is not base 10 and you can’t use your calculator, then the only.

Solving Equations with Logs Day 2. Solving equations with only one logarithm in it: If it is not base 10 and you can’t use your calculator, then the only.

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