Converting Exponential Equations to Logarithmic Equations

16-week Lesson 34 (8-week Lesson 28) Converting Exponential Equations to Logarithmic Equations (Part 2)

As shown in the previous set of notes, exponential equations can be solved by simply converting from exponential form to logarithmic form, and isolating the variable . For example, if 2+4 = 3, I could solve the equation as follows:

2+4 = 3

+ 4 = log2(3) = log2(3) - 4 But not every exponential equation will have an exponential expression (such as 2+4) equal to a non-exponential expression (such as 3). For example, the next equation contains an exponential expression equal to another exponential expression:

2+4 = 35-

This equation can still be solved by converting to logarithmic form: + 4 = log2(35-)

However, in order to bring 5 - out of the exponent to solve for , we need to use the Power Rule for Logarithms.

Power Rule for Logarithms:

log() = log()

A factor multiplied by a logarithm can be re-written as the argument of the logarithm raised to the power of that factor

The Power Rule for Logarithms can also be used to take the exponent of an argument and bring it down to make it a factor.

log() = log()

The exponent of an argument can be re-written as a factor multiplied by the logarithmic expression

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16-week Lesson 34 (8-week Lesson 28) Converting Exponential Equations to Logarithmic Equations (Part 2)

Example 1: Solve the exponential equation 2+4 = 35- by converting to logarithmic form and then isolating the variable . DO NOT APPROXIMATE, ENTER EXACT ANSWERS ONLY.

2+4 = 35-

+ 4 = log2(35-)

+ 4 = (5 - ) log2(3)

+ 4 = 5 log2(3) - log2(3)

Since there are two terms with a factor of ( and log2(3)), I will bring those two terms together on one side of the equation so I can factor out an to solve.

+ log2(3) = 5 log2(3) - 4

(1 + log2(3)) = 5 log2(3) - 4

=

() - + ()

Combining the Power Rule for Logarithms with the idea of converting

from exponential form to logarithmic form from the previous set of notes will allow us to solve exponential equations such as 2+4 = 35-.

When solving these types of equations I usually tend to make the smaller base (2 in this case) the base of the logarithmic expression. I do this because I think it makes logarithmic expressions that can be simplified easier to identify, like we'll see in the next example. However this is not required; the answers are equivalent regardless, and either should be accepted. The only exception to this rule is when the smaller base is a fraction, as we'll see on Example 5a.

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16-week Lesson 34 (8-week Lesson 28) Converting Exponential Equations to Logarithmic Equations (Part 2)

Example 2: Solve the exponential equation 16-1 = 43 by converting to logarithmic form and then isolating the variable . DO NOT APPROXIMATE, ENTER EXACT ANSWERS ONLY.

Once again, keep in mind that when solving exponential equations, we do not have to check our answers like we do with logarithmic equations because there are no restrictions on the domain of exponential equations.

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16-week Lesson 34 (8-week Lesson 28) Converting Exponential Equations to Logarithmic Equations (Part 2)

Example 3: Solve the exponential equation 31-2 = 4+5 by converting to logarithmic form and then isolating the variable . DO NOT APPROXIMATE, ENTER EXACT ANSWERS ONLY.

Keep in mind that solving exponential equations by converting to logarithmic form is not the only possible method. You could also solve these types of equations by using common logarithms or natural logarithms. Doing so will produce answers that will look different than what I come up with by converting to log form, but these answers will still be equivalent; here are a couple of alternate answers for Example 3:

ln(3) - 5 ln(4) ln(4) + 2 ln(3)

log (10324) log(36)

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16-week Lesson 34 (8-week Lesson 28) Converting Exponential Equations to Logarithmic Equations (Part 2)

Example 4: Solve the exponential equation 2-3 = 1- by converting to logarithmic form and then isolating the variable . DO NOT APPROXIMATE, ENTER EXACT ANSWERS ONLY.

2

-

3

=

log ( 1- )

2 - 3 = (1 - ) ln()

2

-

3

=

ln()

-

ln()

- 6 = 2 ln() - 2 ln()

+ 2 ln() = 2 ln() + 6

(1 + 2 ln()) = 2 ln() + 6

() + = + ()

Keep in mind that the equation - 6 = 2 ln() - 2 ln() from the problem above is solved just like any other equation that contains more than one term with a factor of . We need to get our terms together, then solve by factoring. ln() is simply a number that we add, subtract, multiply and divide, just like any other number. The steps used to solve for above are the same steps we've used before when solving other equations.

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