Solving ReviewSemester 2

𝑒𝑥−1=12Notice 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.

𝑙𝑛𝑒𝑥− 1=𝑙𝑛12Because it is now in ln form AND because of the power property of

logs we can move the “x-1” down in front.

(𝑥−1) 𝑙𝑛𝑒❑=𝑙𝑛12

REMEMBER that “ln e” equals 1.

(𝑥−1) 𝑙𝑛𝑒❑=𝑙𝑛12

𝑥−1=𝑙𝑛12Take the ln 12 and add 1.

𝑥=3.48

4𝑒𝑥+ 1=12Notice 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.

𝑙𝑛𝑒𝑥− 1=𝑙𝑛3Because it is now in ln form AND because of the power property of

logs we can move the “x-1” down in front.

(𝑥−1) 𝑙𝑛𝑒❑=𝑙𝑛3

REMEMBER that “ln e” equals 1.

(𝑥−1) 𝑙𝑛𝑒❑=𝑙𝑛3

𝑥−1=𝑙𝑛3Take the ln 3 and add 1.

𝑥=2.1

𝑙𝑜𝑔3 𝑥+4=12Notice 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.

𝑙𝑜𝑔3 𝑥=8Now we MUST rewrite it in exponential form.

38=𝑥𝑥=6561

𝑙𝑜𝑔3 (𝑥+4)=12Notice 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.

𝑥+4=312

𝑥=531,445

𝑙𝑜𝑔3 (𝑥+4)=𝑙𝑜𝑔312Notice we have “log” of the same base appearing on both sides of the

equation. That means we can cancel the logs.

x+4 = 12𝑥=8

𝑙𝑜𝑔3 𝑥+4=𝑙𝑜𝑔312Notice 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.

- 4In order to simplify the

𝑙𝑜𝑔3 𝑥=log 12𝑙𝑜𝑔3

−4

𝑙𝑜𝑔3 𝑥=log 12𝑙𝑜𝑔3

−4We can now key this into our calculator:

log 12 = ÷ log 3 = - 4 =

𝑙𝑜𝑔3 𝑥=−1.74Now we must rewrite it.

𝟑−𝟏.𝟕𝟒=𝒙𝒙=.𝟏𝟒𝟖

) = 5Notice 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.

) = 5Now we must rewrite into exponential form

𝑥 (𝑥+2 )=35

𝑥 (𝑥+2 )=35Foil, set equal to zero and either use

quadratic formula or factor.

𝑥2+2𝑥−243=0

𝑥=−2±√(2)2−4 (243)

2

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

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.

𝑥=−5±√(5)2−4 (8)

2

𝑥=−5±√(5)2−4 (8)

2

Since we get a negative number in the square root sign there is no solution.

We raise both sides to the

(x +2)= 8 x = 6

(𝑥+2 )23=4

=

T

= 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)

4x = 5(x+2)Distribute, collect like terms, set equal to zero and

use factoring or quadratic formula.

4x = 50 =

𝑥=−13±√(13)2−4 (60)

12

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

= 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)

Cancel….

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

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

4(x – 4) = 5 +Distribute, collect like terms, set equal to zero, factor or use quadratic formula.

4x – 16 = 5 +4x – 16 =

0= 𝑥=

−8±√(−8)2−4(10)4

𝑥=−8±√244

𝑜𝑟 𝑥=−4 ±√62

𝑦=𝑥3+8 𝑥2+12Notice 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.

𝑦=𝑥3+8 𝑥2+19𝑥+12Trial # 1: x = 1

1 1 8 19 12

1

1

9

928

28

40Remember that if this

doesn’t result in a 0 then x = 1 IS NOT a factor.

𝑦=𝑥3+8 𝑥2+19𝑥+12Trial # 2: x = -1

-1 1 8 19 12

1

-1

7

-712

-12

0Because the result is a 0

then x = -1 IS a factor.

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

𝑥2+7 𝑥❑+12(x + 3)(x + 4)

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