log notes and exercises

LOGS ARE EXPONENTS!!!! ----LOGS ARE EXPONENTS!!!! The log function is the inverse of the exponential function SO LOGS ARE EXPONENTS!!! Learn to say the log expression properly: log 2 32 is said " log base 2 of 32 ". The 2 common bases for logs are 10 and e (2.718...). we use log x for base 10, ln x for base e. To change from log form to exponential form: base exponent = value. The exponential form of log 2 8 = 3 is 2³ = 8. since 3 is the exponent we apply to base 2 to get 8. The exponential form of log c x = y is c y = x. The base of a log function is a positive Real number. therefore, logs can be negative as in log 7 (1/49) = – 2, but we cannot take the log of negative values. log 7 ( – 49) does not exist. No power of 7 can give you NEGATIVE 49!! log x 1 = 0 because x 0 = 1 Examples: Put these log statements in exponential form. d) log 64 8 = ½ (64) 1/2 = 8 c) log 7 (1/49) = – 2 (7) – 2 = 1/49 b) log 1/2 8 = – 3 (½) – 3 = 8 a) log 3 9 = 2 3 2 = 9 Operations on Logarithmic Expressions: Since logs are exponents, they behave like exponents!!!! Which means that: the log of a product is the sum of the logs, (to multiply, we add exponents) the log of a quotient is the difference of the logs, (to divide, we subtract exponents) the log of an exponentiated term is the product of the log and the exponent. (x ³)² = x 6. Examples: log 2 ( 3xy) = log 2 3 + log 2 x + log 2 y. log 2 ( 3/xy) = log 2 3 – (log 2 x + log 2 y). log 2 (x 5y 3/z 6) = 5 log 2 x + 3 log 2 y – 6 log 2 z. QUESTION If log 2 a = x and log 2 b = y, which expression is equivalent to log 2 8 a2 b b) 3x² y c) 2x – y/2 + 3 d) x² – y 1/2 + 3. 1/2 a) 3x² – y 1/2 . Math 536 Log Notes and ExercisesQUESTIONS 1/Evaluate: c) log (1/3) 3 + 2 log x x d) log 4 (1/64) – log 2 (1/4) + log 95,001(1) 2 – 7 log ( 1/n ) n 4 a) log 2 (8 b) 2 log 100 + 3 log 1000 – log 0.0001 – 1 ) – log 3 9 + log ½ 4 .2/Write as a single log: e) log a 1 + log a a2 + log a a2 d) 5ln x – ½ ln y + 7 ln e – 15 ln e 2c) log 3 27 + . 52 log 3 x − 15 log 3 y b) a log b – 2 log c 2 a) 2 log x + 3 log y – 5 log z + log d . Math 536 Log Notes and ExercisesSOLUTIONS 1/Evaluate: log 4 (1/64) = – 3 since (4) – 3 = 1/64 log 2 (1/4) = – 2 since (2) – 2 = ¼. d) log 4 (1/64) – log 2 (1/4) + log 95,001(1) we have – 3 – ( – 2) + 0 = – 1 log (1/3) 3 = – 1 since (1/3) – 1 = 3 2 log x x 2 = 2 (2 log x x) = 4 (1) = 4 and 7 log ( 1/n ) n 4 = 7(– 4) since (1/n) – 4 = n 4 . c) log (1/3) 3 + 2 log x x 2 – 7 log ( 1/n ) n 4 = so we have 3 + 4 – 7( – 4) = 35 since no base is indicated, base = 10. evaluate the logs and add 'em up. b) 2 log 100 + 3 log 1000 – log 0.0001 = 2 (2) + 3 (3) – (– 4) = 17 apply 3rd rule of logs log x n = n log x to the first term. Evaluate the 2 other logs. Collect the constants. a) log 2 (8 – 1 ) – log 3 9 + log ½ 4 = – log 2 (8) – 2 + (–2) = – 3 – 2 – 2 = – 7 .2/Write as a single log: log a 1 = 0, log a a 2 = 2, and log a (a/2) = log a a – log a 2 = 1 – log a 2 e) log a 1 + log a a2 + log a a2 0 + 2 + 1 – log a 2 = 3 – log a 2 first put the exponents back where they belong. recall that ln e = 1 and ln e 2 = 2 so 7 ln e = 7 and – 15 ln e 2 = – 30 d) 5ln x – ½ ln y + 7 ln e – 15 ln e 2. ln x5 y − 23 first put the exponents back where they belong. now use rules of logs to combine. c) log 3 27 + 52 log 3 x − 15 log 3 y = log 3 27 x5 5 y first put the exponents back where they belong. now use rules of logs to combine. b) a log b – 2 log c 2 + log d = log b a – log c 4 + log d = log [ b a d /c 4 ] first put the exponents back where they belong. now use rules of logs to combine. a) 2 log x + 3 log y – 5 log z = log x ² + log y ³ – log z 5 = log [ x ² y ³/z 5 ] . Math 536 Log Notes and ExercisesSolving Log Equations memory tweak: since the base of any log function is positive (> 0), we can't take the log of negative numbers. Solutions which make the log function argument negative must be dumped. Example: Solve for x. log 2 x + log 2 (x – 2) = 3 apply the rule of logs: log (MN) = log M + log N so log 2 x (x – 2) = 3 Now we change to exponential form 2 3 = x (x – 2) = 8 since 2 3 = 8 x 2 – 2x – 8 = 0 becomes (x – 4)(x + 2) = 0 so x = 4 or x = –2 . We can't take log 2 (–2) since there is no power of 2 that gives us –2. The solution is x = 4. To solve a log equation: using rules of logs, we combine the terms into a single log term set that log term = k ; some constant change to exponential form, then solve for the unknown. Note: watch for logs you can evaluate before solving like this Example 4 log 2 x + log 34 64 27 = 1 , but we know that log 34 64 27 = −3 so log 2 x − 3 = 1 t log 2 x = 4, therefore 2 4 = x = 4 Sometimes, we have to solve for an unknown base, like this Example 5 log x 18 + 5 = 2 t log x 18 = − 3, so x − 3 = 18 x − 3 = 1 x3 , so, 1 x3 = 18 t x = 2 Math 536 Log Notes and Exercises3/Solve for x. Identify any restrictions : g) log x (1 /9) – log x 1 = log x 27 – 5 e) log 6 (x + 3) + log 6 (x – 2) = 1 f) 3 log (0.5x + 2) 2 – 1 = 5 c) log 3 x + log 3 (x + 6) = 3 d) log 3 x+ log 3 2 = 4 a) log 2 (2x – 1) + log 2 (x – 1) = 0 b) log 4 (x – 3) – log 8 64 = log 4 (x + 1) Math 536 Log Notes and Exercises3/Solve for x: Combine the logs first, put it in exponential form solve the quadratic. x = – 9 is not a solution (negative values) c) log 3 x + log 3 (x + 6) = 3 log 3 [ x (x + 6)] = 3 so x (x + 6) = 3 ³ = 27. we solve x ² + 6x – 27 = 0 we get (x + 9)(x – 3) = 0 so x = 3. rewrite log 8 64 as 2, transpose the terms then same deal as a). b) log 4 (x – 3) – log 4 (x + 1) = 2 now log4 x − 3 x + 1 = 2 e x − 3 x + 1 = 42 = 16 so, x – 3 = 16x + 16 or x = – 19 /15 Combine the logs first, put it in exponential form solve the quadratic. but x = 0 makes both logs undefined a) log 2 (2x – 1) + log 2 (x – 1) = 0 so log 2 (2x – 1)(x – 1) = 0 which means that 2x² – 3x + 1 = 2 0 = 1 so, 2x² – 3x = 0 or x(2x – 3) = 0 solutions are x = 0 or x = 1.5 solution is x = 1.5. sum of logs = log of the product change to exponential form, solve. e) log 6 (x + 3) + log 6 (x – 2) = 1 log 6 [(x + 3)(x – 2)] = 1 Now 6 1 = (x + 3)(x – 2), which becomes x 2 + x – 12 = 0 so (x – 3)(x + 4) and x = 3 combine sum of logs = log of the product change to exponential form, solve. d) log 3 x + log 3 2 = 4 log3 2x = 4 becomes 3 4 = 2x x = 81 /2 Math 536 Log Notes and Exerciseslogx 1 = 0, transpose log x 27 combine logs by division, exponential form solve. g) log x (1 /9) – log x 1 = log x 27 – 5 log x (1 /9) – log x 27 = – 5 log x 19 27 = − 5 t x− 5 = 1 32 $ 33 = 3− 5 so x = 3 transpose the ( – 1), divide by 3 change to exponential form, take square root transpose +2, and solve. f) 3 log (0.5x + 2) 2 – 1 = 5 log (0.5x + 2) 2 = 2 (0.5x + 2) 2 = 10 2, so 0.5x + 2 = 10 this makes ½x = 8, so x = 16 . 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