Properties of Logarithms: Solutions

Here are some solutions to the Exercises to accompany the section Properties of Logarithms.

Two Special Logarithms

Use your calculator to compute the following values. Check your answer by plugging into the corresponding exponential function.

log(8213)

Solution

Notice that, since this is close to 8213 » 10,000, we would expect the value to be close to 4. Our calculator gives log(8213) » 3.9145.
log(0.05217)

Solution

Since 0.1 > 0.05217 > .01, we expect the value to be between -1 and -2, and, in fact, our calculator gives log(0.05217) » -1.28258.
ln(4.148)

Solution

Since 4.148 > e » 2.71828..., we expect the value to be greater than 1. Our calculator gives ln(4.148) » 1.42263.
ln(1.321)

Solution

Since 1 < 1.321 < e » 2.71828..., we expect the value to be between 0 and 1. Our calculator gives ln(1.321) » 0.278389.
ln(0.215)

Solution

Since 0 < 0.215 < 1, we expect the value to be negative. Our calculator gives ln(0.215) » -1.53712
log(-275)

Solution

From the graph, we know that the domain is {x | x > 0}, so that log(-275) is undefined.

Note: You can use XFunctions to check your answers, as well, by using the "x = " box:

Back to Exercises.

Laws of Exponents and Logarithms

Use your calculator, and the change-of-base formula, with either the natural or common logarithm, to compute the following values.

log_0.75(4)

Solution

Using the natural logarithm gives log_0.75(4) = ln(4)/ln(0.75) » 1.38629/(-0.287682) » -4.81884. Using the common logarithm, we have log_0.75(4) = log(4)/log(0.75) » 0.60206/(-0.124939) » -4.81884. This answer agrees, to three digits, with the answer we obtained before of -4.82.
log_0.5(0.35)

Solution

Using the natural logarithm gives log_0.5(0.35) = ln(0.35)/ln(0.5) » (-1.04982)/(-0.693147) » 1.51457. Using the common logarithm, we have log_0.5(0.35) = log(0.35)/log(0.5) » (-0.455932)/(-0.30103) » 1.51457. This answer agrees, to three digits, with the answer we obtained before of 1.51.
log_0.49(0.343)

Solution

Using the natural logarithm gives log_0.49(0.343) = ln(0.343)/ln(0.49) » (-1.07002)/(-0.71335) » 1.5. Using the common logarithm, we have log_0.49(0.343) = log(0.343)/log(0.49) » (-0.464706)/(-0.309804) » 1.5. This answer agrees, to three digits, with the answer we obtained before of 1.5.

You should notice how cool it is that, in both cases, two horrendous (irrational) numbers simplify down to a very simple decimal. However, since we used a calculator, we could not be sure that this value is exactly 1.5, unless we recognize both 0.49 and 0.343 as powers of 0.7 (i.e., (0.7)² = 0.49 and (0.7)³ = 0.343)! We can then use the change-of-base formula with base 0.7 to show that log_0.49(0.343) = log_0.7(0.343)/log_0.7(0.49) = 3/2 = 1.5.
log₂(5)

Solution

Using the natural logarithm gives log₂(5) = ln(5)/ln(2) » (1.60944,)/(0.693147) » 2.32193. Using the common logarithm, we have log_0.75(4) = log(5)/log(2) » (0.69897)/(0.30103) » 2.32193. Notice that this answer does not agree, to three digits, with the answer we obtained before of 2.33. This is due to round-off error in XFunctions, since we actually used that program to estimate the solution to -14 = -3·2^x + 1.
Fill-in-the-blanks to create your own Exercise: log_{_}(__).

Solution

Check your answer by plugging into the corresponding exponential function. For example, if you created the Exercise to calculate log_1.41(3.14) and used the change-of-base formula to obtain a value of 3.3302, then you would plug into exp_1.41 to obtain exp_1.41(3.14) = 1.41^3.3302 » 3.14.

Repeat this Exercise as often as necessary until you are confident in your ability to use the change-of-base formula.

Back to Exercises.

Solving Exponential Equations

For each of the following pairs of equations, either:

Determine which rule of logarithms was used to go from the first equation to the second, and how it was used (i.e., match up the variables in the formula for the rule and the corresponding expressions in the equation), or
Explain how the step misused a rule of logarithms.

-7 = 4 - 5 log₃(3^x) Þ -7 = 4 - 5x

Solution

This uses the inverse property of logarithms, "log_b(b^x) = x", with b « 3 (and x « x), to replace log₃(3^x) by x.
-7 = 4 - 5 log₃(2^x) Þ -7 = 4 - 5x

Solution

This is a misuse of the inverse property of logarithms, "log_b(b^x) = x", since the base of the logarithm is 3, but the base of the exponential is 2..
-7 = 4 - 5 log₃(2^x) Þ -7 = 4 - 5x log₃(2)

Solution

This uses the power property of logarithms, "log_b(s^t) = t log_b(s)", with b « 3, s « 2, and t « x), to replace log₃(2^x) by x log₃(2).
3 = log₂(5·2^x) Þ 3 = x log₂(5·2)

Solution

This is a misuse of the power property of logarithms, since what is in the logarithm is a product, not simply a power.
3 = log₂(5·2^x) Þ 3 = log₂(5) + log₂(2^x)

Solution

This uses the product property of logarithms, "log_b(st) = log_b(s) + log_b(t)", with b « 2, s « 5, and t « 2^x), to replace log₂(5·2^x) by log₂(5) + log₂(2^x).
2 = log₇(5/x) Þ 2 = log₇(5)/ log₇(x)

Solution

This is a misuse of the quotient property of logarithms, since it says that "Logarithms turn quotients into differences". In fact, by the change-of-base formula, log₇(5)/ log₇(x) = log_x(5), which is quite different than log₇(5/x); you can see this by comparing their graphs in the Example of the following applet:
2 = 4 log₇(5/x) Þ 2 = 4 log₇(5) - log₇(x)

Solution

Although this correctly uses the quotient property of logarithms, "log_b(s/t) = log_b(s) - log_b(t)", with b « 7, s « 5, and t « x), to replace log₇(5/x) by log₇(5) - log₇(x), it is ultimately incorrect, because it fails to use proper parentheses when substituting. The correct result is "4(log₇(5) - log₇(x))". You can see the difference between these two results by comparing their graphs in the Example of the following applet:
Have your partner create a similar Exercise, by either correctly or incorrectly applying a rule of logarithms to some expression.

Solution

Your partner should correct your work.

Repeat this Exercise as often as necessary until you are confident in your ability to correctly apply the rules of logarithms.

Back to Exercises.

Use algebra and rules of logarithms to solve the following exponential equations. Give your answer to at least 6 correct digits. Check your answer by plugging back into the equation.

191 = 3^x/9.72 - 416

Solution: Working "outside-in", focusing on the right side of the equation, we obtain:

Plugging this back into the equation gives:
3^7.90335/9.72 - 416 » 191.002,
which is very close to 191.

Equation	Reason
191 + 416 = 3^x/9.72 - 416 + 416	Add 416 to both sides.
607 = 3^x/9.72	Simplify.
9.72·607 = 9.72·3^x/9.72	Multiply both sides by 9.72.
5900.04 = 3^x	Simplify.
ln(5900.04) = ln(3^x)	Apply natural log to both sides.
ln(5900.04) = xln(3)	Use the power rule.
ln(5900.04)/ln(3) = x	Divide both sides by 9.72.
x » 8.68271/1.09861 » 7.90335	Simplify.

3^{(x + 1)} = 26·2^-x

Solution: If we proceed as in the similar example of the text, we would:
Plugging this back into the left side of the equation gives:
3⁽^1.20523^{+ 1)} » 11.2762,
while the right side:
26·2^-^1.20523 » 11.2762,
gives the same result to 6 decimal places.
If we apply the product and quotient rule to our answer, x = (ln(26) - ln(3))/(ln(3) + ln(2)) = ln(26/3)/ln(3·2) = ln(26/3)/ln(6), this suggests an alternative approach to solving this equation:
In general, you can often avoid the explicit use of rules of logarithms by using rules of exponents, and vice versa. Which approach to take is largely a matter of which rules with which you are more comfortable and which way seems more natural to you.

Equation	Reason
ln(3^{(x + 1)}) = ln(26·2^-x)	Apply natural log to both sides.
(x + 1)ln(3) = ln(26) + ln(2^-x)	Use the power rule on the left and the product rule on the right.
ln(3)x + ln(3) = ln(26) + (-x)ln(2)	Distribute on the left and use the power rule on the right.
ln(3)x + ln(2)x = ln(26) - ln(3)	Add ln(2)x to and subtract ln(3) from both sides.
(ln(3) + ln(2))x = ln(26) - ln(3)	Factor out x.
x = (ln(26) - ln(3))/(ln(3) + ln(2)) » 1.20523	Divide both sides by the coefficient of x.

Equation	Reason
3^{(x + 1)}/2^-x = 26	Divide both sides by 2^-x.
3·3^x2^x = 26	Use the product rule of exponents on the left and the reciprocal rule of exponents in the denominator.
3·3^x2^x/3 = 26/3	Divide both sides by 3..
3^x2^x = 26/3	Simplify.
(3·2)^x = 26/3	Use the distributive rule of exponents on the left and the reciprocal rule of exponents in the denominator.
6^x = 26/3	Simplify.
log₆(6^x) = log₆(26/3)	Apply log₆ to both sides.
x = ln(26/3)/ln(6)	Use the inverse rule on the left and the on the right.

3·2^{(x - 2)/7} - 35 = 1501

Solution: This is similar to part a), except we must focus on the left side of the equation:

Plugging this back into the equation gives:
3·2^{(65 - 2)/7} - 35 = 3·2⁹ - 35 = 3·512 - 35 = 1501.
Note: If we had recognized 512 as a power of 2 earlier, we could have taken log₂ at the 5th step, obtained (x - 2)/7 = 9, and the exact answer of x = 65.

Equation	Reason
3·2^{(x - 2)/7} - 35 + 35 = 1501 + 35	Add 35 to both sides.
3·2^{(x - 2)/7} = 1536	Simplify.
3·2^{(x - 2)/7}/3 = 1536/3	Divide both sides by 3.
2^{(x - 2)/7} = 512	Simplify.
ln(2^{(x - 2)/7}) = ln(512)	Apply natural log to both sides.
((x - 2)/7)ln(2) = ln(512)	Use the power rule.
x - 2 = 7ln(512)/ln(2)	Divide both sides by ln(2), multiply by 7, and simplify.
x - 2 + 2 = 7ln(512)/ln(2) + 2	Add 2 to both sides.
x » 65.	Simplify.

21·2^{(x + 1)}3^{(1 - x)} = 56

Solution: This is similar to part b):
Plugging this back into the left side of the equation gives:
21·2^{(2 + 1)}3^{(1 - 2)} = 21·8·(1/3) = 56.
Again, if we use rules of logarithms, we can rewrite the answer:
x = (ln(56) - ln(21) - ln(2) - ln(3))/(ln(2) - ln(3)) = ln(56/(21·2·3))/ln(2/3) = ln(4/9)/ln(2/3) = log_2/3(4/9),
Since (2/3)² = 4/9, we can be sure that the answer is exactly 2.

Have your partner create a similar Exercise.

Solution

Your partner should correct your work.

Repeat this Exercise as often as necessary until you are confident in your ability to correctly apply the rules of logarithms to solve exponential equations.

Equation	Reason
ln(21·2^{(x + 1)}3^{(1 - x)}) = ln(56)	Apply natural log to both sides.
ln(21) + (x + 1)ln(2) + (1 - x)ln(3) = ln(56)	Use the product and then the power rules.
ln(21) + ln(2) + ln(3) + ln(2)x - ln(3)x = ln(56)	Distribute and collect like terms.
ln(2)x - ln(3)x = ln(56) - ln(21) - ln(2) - ln(3)	Subtract constant terms.
(ln(2) - ln(3))x = ln(56) - ln(21) - ln(2) - ln(3)	Factor out x.
x = (ln(56) - ln(21) - ln(2) - ln(3))/(ln(2) - ln(3)) » 2.	Divide both sides by the coefficient of x and simplify.

Back to Exercises.