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Solving Review Semester 2. Notice the variable is in the exponent. That means we need to use logs to solve. Because an “e” is involved we must use ln. Take the ln of both sides. Because it is now in ln form AND because of the power property of logs we can move the “x-1” down in front.

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Presentation Transcript
slide2

Notice the variable is in the exponent. That means we need to use logs to solve.

Because an “e” is involved we must use ln.

Take the ln of both sides.

Because it is now in ln form AND because of the power property of logs we can move the “x-1” down in front.

slide3

REMEMBER that “ln e” equals 1.

Take the ln 12 and add 1.

slide4

Notice the variable is in the exponent. That means we need to use logs to solve.

Same as last one, but MUST divide both sides by 4 first.

Because an “e” is involved we must use ln.

Take the ln of both sides.

Because it is now in ln form AND because of the power property of logs we can move the “x-1” down in front.

slide5

REMEMBER that “ln e” equals 1.

Take the ln 3 and add 1.

slide6

Notice we have “log” appearing in the equation. That means we need to use exponents to solve.

Because the x and 4 are NOT grouped together in parentheses we must subtract 4 from both sides first.

Now we MUST rewrite it in exponential form.

slide7

Notice we have “log” appearing in the equation. That means we need to use exponents to solve.

Because the x and 4 are grouped together we must rewrite it in exponential form.

slide8

Notice we have “log” of the same base appearing on both sides of the equation. That means we can cancel the logs.

x+4 = 12

slide9

Notice we have “log” of the same base appearing on both sides of the equation BUT the 4 is NOT grouped with the log so we must deal with it BEFORE we rewrite.

- 4

In order to simplify the

slide10

We can now key this into our calculator:

log 12 = ÷ log 3 = - 4 =

Now we must rewrite it.

slide11

) = 5

Notice we have “log” of the same base appearing on the SAME side of the equation.

WE CANNOT CANCEL THE LOGS!

We must instead rewrite the equation in terms of a single log using the log properties.

) = 5

Now we must rewrite into exponential form

slide12

Foil, set equal to zero and either use quadratic formula or factor.

Since we get a negative under the square root there is no solution.

slide13

Square both sides to get rid of the square root. Don’t forget to FOIL the “ x – 2”

After moving the x and -4 over we either factor or use quadratic formula.

slide15

We raise both sides to the

=

T

(x +2)= 8

x = 6

slide16

=

Eliminate the denominators by multiplying each of the three terms by the lowest common denominator which is x(x+2)

x(x+2) = x(x+2)

Cancel….

x(x+2) = x(x+2)

4x = 5(x+2)

slide17

4x = 5(x+2)

Distribute, collect like terms, set equal to zero and use factoring or quadratic formula.

4x = 5

0 =

Since we get a negative under the square root there is no solution.

slide18

=

Eliminate the denominators by multiplying each of the three terms by the lowest common denominator BUT to find the LCD we must factor FIRST, (x+2)(x-4). The LCD is then (x+2)(x-4)

(x -4)(x+2) = (x-4)(x+2)

Cancel….

(x -4)(x+2) = (x-4)(x+2)

slide19

4(x – 4) = 5 +

Distribute, collect like terms, set equal to zero, factor or use quadratic formula.

4x – 16 = 5 +

4x – 16 =

0=

slide20

Notice that this is a cubic. You hope that you can solve it by taking out an x from each term but you cannot because the last term has no x.

You hope that it has FOUR terms because then you might be able to solve by grouping the first two factors and the last two factors and solve by factoring. You cannot since there are only 3 terms.

You have to find the p’s and q’s technique. Remember that the coefficient of the cubed term is the q and the 12 is the p term.

Step 1: Factor 12. Remember that we have to include the negative and positive factors.

± 1, ±2, ±3, ±4, ±6, ±12

Step 2: Now using synthetic division we see which factor divides evenly into the polynomial.

slide21

Trial # 1: x = 1

1

1

8

19

12

28

1

9

Remember that if this doesn’t result in a 0 then x = 1 IS NOT a factor.

28

1

9

40

slide22

Trial # 2: x = -1

-1

1

8

19

12

-12

-1

-7

Because the result is a 0 then x = -1 IS a factor.

12

1

7

0

Now we rewrite into a quadratic and either solve using quadratic formula or factoring.

(x + 3)(x + 4)

That means the solutions are: -1,-3, -4