Exponential & Logarithmic Equations

[Pages:5]Exponential & Logarithmic Equations

This chapter is about using the inverses of exponentials or logarithms to solve equations involving exponentials or logarithms.

Solving exponential equations

An exponential equation is an equation that has an unknown quantity, usually called x, written somewhere in the exponent of some positive number. Here are three examples of exponential equations: x = 5, or 23x 5 = 2, or

e 35x 1 = 3x. In all three of these examples, there is an unknown quantity, ,

x that appears as an exponent, or as some part of an exponent.

To solve an exponential equation whose unknown quantity is , the first step x

is to make the equation look like f(x) = where ( ) is some function, and

ac

fx

a

and c are numbers. Sometimes the equation will already be set up to look like

this, as in the first two examples above of x = 5 or 23x 5 = 2. Sometimes, e

you'll have to use the rules of exponentials to make your equation look like

f(x) = . In the third example from the previous paragraph, we could divide ac the equation 35x 1 = 3x by 3x to obtain 35x 1 x = 1, which is the same thing

as 34x 1 = 1. (In this last sentence we used the rule of exponentials that x a

divided by y equals x y.) Now all three of our exponential equations have

a

a

the form f(x) = . ac

Once your equation looks like f(x) = , you can erase the exponential

a

c

base on the left side of the equation by applying its inverse function, log , a

a

to the right side of the equation. That would leave you with the equation

( ) = log ( ). Sometimes you can write the number log ( ) as a more

fx

c

c

a

a

familiar number. Sometimes you can't. Either way, it's just a number

If x = 5, then = log (5). If 23x 5 = 2, then 3 5 = log (2) = 1. If

e

x

x

e

2

34x 1 = 1, then 4 1 = log (1) = 0.

x

3

At this point in the problem, you might already be finished. If not, you

should be able to solve for x using techniques that we've learned or reviewed earlier in the semester. In the three examples above, the answers would be

= log (5), = 2, and = 1 respectively.

x

ex

x

4

log (5) is a perfectly good number. Just as good as say 17, or 2. There's

no

e

way

to

simplify

it.

You

should

be

comfortable

with

it

as

an

ans3wer.

222

Steps for solving exponential equations

Step

1:

Make

the

equation

look

like

f (x)

a

=

c

where

a, c

2

R

and

() fx

is

a

function.

Step 2: Rewrite the equation as ( ) = log ( ).

fx

ac

Step 3: Solve for . x

Example. Let's solve for if x

3x 7 = 5 x 1

e

e

To perform Step 1, we can divide both sides of the equation by x 1. We'd e

be left with

3x 7

e =5

x1

e

But

3x 7

e

x1

= 3x e

7

(x

1) = 2x e

6.

So we're really left with

e

2x 6 = 5 e

and that completes Step 1.

Step 2 is to erase the exponential function in base from the left side of e

the equation 2x 6 = 5 by applying its inverse, the logarithm base , to the

e

e

right side of the equation. To put it more simply, we rewrite 2x 6 = 5 as e

2 6 = log (5)

x

e

Step 3 is to solve the equation 2 6 = log (5) using algebra. We can do x

e

this by adding 6 and then dividing by 2. We'll be left with the answer

log (5) + 6

=e

x

2

*************

223

Solving logarithmic equations

A logarithmic equation is an equation that contains an unknown quantity,

usuallpy called

, x

inside

of

a

logarithm.

For example, log (5 ) = 3, and x

2

log ( ) = 1, and log ( 2) = 7 log (2 ) are all logarithmic equations.

x

x

x

10

e

e

To solve a logarithmic equation for an unknown quantity , you'll want to x

put

your

equation

into

the

form

log (

a

f (x)

)

=

c

where

f (x)

is

a

functpion

of

x

and is a number. The logarithmic equations log (5 ) = 3 and log ( ) = 1

c

x

x

2

10

are already written in the form log ( ( ) ) = , but log ( 2) = 7 log (2 )

fx c

x

x

a

e

e

isn't. To arrange the latter equality into our desired form, we can use rules

of logarithms. More precisely, add log (2 ) to the equation and use the

x

e

logarithm rule that log ( 2) + log (2 ) = log ( 22 ). Then the equation

x

x

xx

e

e

e

becomes log (2 3) = 7, and that's the form we want our logarithmic equations

x

e

to be in.

Once your equation looks like log ( ( ) ) = , use that the base expo-

fx c

a

a

nential is the inverse of log to rewrite your equation as ( ) = c. You fx a

a

might want to simplify the number that appears as c in your new equation, a

but other than that, you're done with exponentials and logarithms at this

point in the problem. It's time to solve the equation using techniques we

used earlier in the semester.

Let's look at the three examples above. We would rewrite log (5 ) = 3 as

x

2

5 = 23 = 8. Then we solve our new equation to find that = 8.

x

x

5

p

p

We would rewrite log ( ) = 1 as = 101 = 10. Since squaring is the

x

x

10

inverse of the square root, we are left with = 102 = 100.

x

For

the

third

equation,

we

hadqloge(2x3)

=

7.

Rewrite

it

as

23 x

=

7, e

and

then solve for

to find that

=

3

7

e

.

x

x

2

224

Steps for solving logarithmic equations

Step

1:

Make

the

equation

look

like

log

( ( )) fx

=

c

where

a, c

2

R

and

() fx

a

is a function.

Step 2: Rewrite the equation as ( ) = c. fx a

Step 3: Solve for . x

Example. Let's solve for if x

log ( 2 + 2 ) = log ( ) + 4

xx

x

e

e

To perform Step 1, we can subtract log ( ) from both sides of the equation x

e

to get

log ( 2 + 2 ) log ( ) = 4

xx

x

e

e

Recall that log (

2+2 )

log (

) = log (

2

x +2x

)

=

log

(

+ 2). That

xx

x

x

e

means that

e

ex

e

That's the end of Step 1.

log ( + 2) = 4 x

e

Step 2 is to erase the logarithm base from the left side of the equation e

log ( + 2) = 4 by applying the exponential function of base to the right

x

e

e

side of the equation. That is, we rewrite log ( + 2) = 4 as x

e

+2= 4

x

e

Step 3 is to solve the equation + 2 = 4 using algebra. Subtracting 2

x

e

and multiplying by 1 leaves us with the answer

=2 4

x

e

*************

225

Exercises

Solve the following exponential equations for x. 1.) 103x = 1000

2.) 6(14x) = 30

3.) 2 x = 8 e

4.) x + 10 = 17 e

5.) (3x)5 = 27

6.) 5

= x

21

5

7.) 53x 4 = 125

2

8.)

2x

e

=

x

e

2

e

9.)

2

x

=

x+5

11

e ee

Solve the following logarithmic equations for x.

10.) log ( 5) = 2 x

3

11.) log ( ) = 6 x

e

12.) log (2 ) = 24

x

e

p

13.) log (

4) = 5

x

e

14.) log ( 7) = 28 x

2

15.) log (( + 1) 5) = 15 x

10

16.) 5 + log ( 3) = 11 x

e

17.) log ( ) log ( 4) = 3

x

x

2

2

18.) log ( 2) = 3 x

2

226

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