Math 536 Log Exercise

Remember that logs are exponents so they behave like exponents!!!!

That means that a product is a sum, a quotient is a difference, and
exponentiation becomes a product.

Also remember, when evaluating a log, you're looking for the exponent to which
we raise the base to get the number. So log 2 32 = 5 because 2 5 = 32.

Last Note: Learn to say the log expression properly. log 2 32 is said " log base 2 of 32 ".

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QUESTIONS

1/ Evaluate:

 a) log 2 (8 – 1 ) – log 3 9 + log ½ 4 = b) 2 log 100 + 3 log 1000 – log 0.0001 = c) log (1/ 3) 3 + 2 log x x 2 – 7 log ( 1/ n ) n 4 =

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2/ Write as a single log:

 a) 2 log x + 3 log y – 5 log z b) a log b – 2 log c 2 + log d

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3/ Solve for x:

 a) log 2 (2x – 1) + log 2 (x – 1) = 0 b) log 4 (x – 3) – log 8 64 = log 4 (x + 1) c) log 3 x + log 3 (x + 6) = 3

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SOLUTIONS

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1/ Evaluate:

 a) log 2 (8 – 1 ) – log 3 9 + log ½ 4 = – log 2 (8) – 2 + (–2) = – 3 – 2 – 2 = – 7 apply 3rd rule of logs log x n = n log x to the first term. Evaluate the 2 other logs. Collect the constants. b) 2 log 100 + 3 log 1000 – log 0.0001 = 2 (2) + 3 (3) – (– 4) = 17 since no base is indicated, base = 10. evaluate the logs and add 'em up. c) log (1/ 3) 3 + 2 log x x 2 – 7 log ( 1/ n ) n 4 = so we have 3 + 4 – 7( – 4) = 35 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 .

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2/ Write as a single log:

 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 ] = 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.

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3/ Solve for x:

 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. Combine the logs first, put it in exponential form solve the quadratic.but x = 0 makes both logs undefined b) log 4 (x – 3) – log 4 (x + 1) = 2now so, x – 3 = 16x + 16 or x = – 19 / 15 rewrite log 8 64 as 2, transpose the terms then same deal as a). 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. Combine the logs first, put it in exponential form solve the quadratic.x = – 9 is not a solution (negative values)

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