solving equations involving logarithmic and exponential functions on completion of this module you...
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Solving Equations Involving Logarithmic and Exponential Functions
On completion of this module you will be able to: convert logarithmic with bases other than 10 or e use the inverse property of exponential and logarithmic functions to simplify equations understand the properties of logarithms use the properties of logarithms to simplify equations solve exponential and logarithmic equations
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Bases other than 10 or e
Most calculators have log x (base 10) and ln x (base e). How can we solve equations involving bases other than 10 or e? One way is using the change of base rule:
log ln log or log
log lnb b
x xx x
b b
2
6
5
4
2
1
3
1. log 40
2. log 25
3. log 16
14. log
165. log 9
Example
3
6
log 40 ln 401. log 40 or
log6 ln 6
2.0588 to 4 decimal places
Answer
25
log 252. log 25 =2 note that 5 =25
log5
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log163. log 16 =2 note that 4 =16
log 4
4
42
1log1 1164. log 4 note that 2 =
16 log 2 16
Answer
1
3
log95. log 9 2
1log
3
5
Using the inverse property
When an exponential function and a logarithmic function have the same base, they are inverses and so effectively cancel each other out.
ExampleSolve for x: log 1.5.x
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We can’t divide by log! Use the exponential function with the same base (10) – called taking the anti-log. The left and right sides of the equation become exponents with a base of 10
log 1.510 10x
log 1.5x
1.510 31.6228x
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ExampleSolve for x:AnswerQuick solution is to rearrange using the definition of logs: Alternative:
Now rearrange to isolate the variable:
3log 4.x
43log 4 3 81.x x x
3
loglog 4 4
log3
xx
log 4log3x
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Take anti-logs:
log 4log3x
log 4log3
4log3
10 10
10 81
x
x
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ExampleSolve:AnswerWe have an exponential function (base e) which we can cancel out by taking the logarithm with the same base (ln x):
3 2.xe
3 2xe
3ln ln 2xe 3 ln 2x
ln 20.231049 to 6 decimal places
3x
10
log log logmn m n
Rule 1
Properties of logarithms
Example 1
(Since 84 = 12 7)
log84 log12 log7
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log log logmn m n
Rule 1
Example 2Solve for x:Answer
log log 2 2x x
log log 2 2 log 2 2x x x x
2log 2 2x 2log 2 210 10
x
22 100x 2 50x
7.0711 to 4 decimal placesx 12
Note that although both +7.0711 and −7.0711 square to give 50, only +7.0711 solves the original equation.Check:
as required, but
is undefined.Always check that your answer solves the original problem!!
log 7.0711... log 2 7.0711 2
log 7.0711... log 2 7.0711
log log 2 2x x
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log log logm
m nn
Rule 2
Example9
log log9 log66
0.9542425 0.7781512
0.1760912
9log log1.5
6
0.1760912
or
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log logrm r m
Rule 3
Example
log 3 6x
log 3 6x
6
12.5754log 3
x
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1log log m
m
Rule 4
Example
1log log5 0.69897
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Note: Rules 1 to 4 have been expressed in base 10, but are equally valid using any base.
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log1 0 and ln1 0 Rule 5
This rule also works for any base e.g.
since
5 12log 1 0 and log 1=0
0 05 1 and 12 1
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log10 1 and ln 1 e Rule 6
Rule 6 also extends to other bases. Whenever we take the log of the same number as the base, then the answer is 1.e.g. 3 42 10,000log 3 1, log 42 1, log 10,000 1, etc
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log 10 and ln r rr e r
Rule 7
Let’s use Rules 3 and 6 to show why Rule 7 is true.
log 10 log10 1r r r r
ln ln 1re r e r r
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log ln 10 and x xx e x
Rule 8
This uses the concept of log and exponential functions as inverses as we discussed earlier. This rule also works for other bases.
82 loglog2 , 8 , etcxx x x
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log ln log
log lnb
x xx
b b
Rule 9
Recall that this is the change of base formula used earlier.
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Example
log log 2 1x
log 2 1x
log 2 110 10x
2 10x 5x
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Example
34 7x
3log 4 log7x
3 log 4 log7x
log73
log 4x
log73 1.5963
log 4x
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Example3 17 2 1xe 3 17 3xe
3 1 3
7xe
3 1 3ln ln
7xe
33 1 ln
7x
33 ln 1
7x
3ln 1
7 0.05093
x
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1. log log logmn m n
2. log log logm
m nn
3. log logrm r m 1
4. log log mm
5. log1 0 and ln1 0 6. log10 1 and ln 1e
7. log 10 and lnr rr e r log ln8. 10 and x xx e x
log ln9. log
log lnb
x xx
b b
Summary: Rules of Logarithms
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In solving equations which involve exponential and logarithmic terms, the following properties allow us to remove such terms and so simplify the equation.
. then, If
. then , log log If
nmbb
nmnmnm
Exponential and logarithmic equations
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Example
log 4 1 log 2x x
4 1 2x x
3 1x 1
3x
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Example
1255 xx
)1(255 xx
22 xx
2 x
2x
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Example
It doesn’t matter whether base 10 or base e is used, the result will be the same.
8.5 12x
log 8.5 log12x
log8.5 log12x log12
log8.5
1.0791812
0.92941891.1611354
x
ln 8.5 ln12x
ln8.5 ln12x ln12
ln8.52.4849066
2.14006621.1611354
x
The numbers are different but the result is the same.
Base 10 Base e
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Example
The demand equation for a consumer product is
Solve for p and express your answer in terms of common logarithms. Evaluate p to two decimal places when q = 60.
80 2 .pq
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pq 280 qp 802
)80log()2log( qp
)80log(2log qp
2log
)80log( qp
Answer
31
2log
)6080log(
,60When
p
q
3219.4
places) decimal 2 (to
32.4
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Suppose that the daily output of units of a new product on the tth day of a production run is given by:
Such an equation is called a learning equation and indicates as time progresses, output per day will increase.
This may be due to a gain in a worker’s proficiency at his or her job.
Example
0.2500 1 tq e
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Determine, to the nearest complete unit, the output on (a) the first day and (b) the tenth day after the start of a production run.
(c) After how many days will a daily production run of 400 units be reached? Give your answer to the nearest day.
Example (continued)
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a) On the first day or production, t =1, so the daily output will be
b) When t =10,
Note that since the answers to parts (a) and (b) are the number of units of a new product, we have rounded these to the nearest whole unit.
Answer 0.2500 1 tq e
0.2 1500 1 90.6346 91 units.q e
0.2 10500 1 432.3324 432 units.q e
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c) The production run will reach 400 units when q = 400 or at
0.2500 1 tq e
0.2400 500 1 te 0.2400
1500
te
0.2 4 11
5 5te
0.2 1ln ln
5te
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10.2 ln
5t
1ln
5 8.0472 9 days0.2
t
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Notice that the question requires the answer to be rounded to the nearest whole day.
If the answer were round to 8 days,
so production has not quite reached 400.
For this reason we round the answer to 9 days, even though production will be well passed 400 by the end of the 9th day.
0.2 8500 1 399.0517 unitsq e
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As always, we must check that the mathematically obtained solution answers the original question.
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