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 ".
.
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 = |
.
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 |
.
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 |
.
.
SOLUTIONS
.
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 . |
.
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. |
.
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) |
.
.
(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - 200?).