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) |
.
Back toMathTub Index
.
(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - 200?).