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) = 2 now 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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