Download - Solving Review Semester 2
Solving ReviewSemester 2
ππ₯β1=12Notice the variable is in the exponent. That means we need to use
logs to solve.
Because an βeβ is involved we must use ln.Take the ln of both sides.
ππππ₯β 1=ππ12Because it is now in ln form AND because of the power property of
logs we can move the βx-1β down in front.
(π₯β1) πππβ=ππ12
REMEMBER that βln eβ equals 1.
(π₯β1) πππβ=ππ12
π₯β1=ππ12Take the ln 12 and add 1.
π₯=3.48
4ππ₯+ 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 βeβ is involved we must use ln.Take the ln of both sides.
ππππ₯β 1=ππ3Because it is now in ln form AND because of the power property of
logs we can move the βx-1β down in front.
(π₯β1) πππβ=ππ3
REMEMBER that βln eβ equals 1.
(π₯β1) πππβ=ππ3
π₯β1=ππ3Take the ln 3 and add 1.
π₯=2.1
πππ3 π₯+4=12Notice we have βlogβ 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.
πππ3 π₯=8Now we MUST rewrite it in exponential form.
38=π₯π₯=6561
πππ3 (π₯+4)=12Notice we have βlogβ 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.
π₯+4=312
π₯=531,445
πππ3 (π₯+4)=πππ312Notice we have βlogβ of the same base appearing on both sides of the
equation. That means we can cancel the logs.
x+4 = 12π₯=8
πππ3 π₯+4=πππ312Notice we have βlogβ 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
πππ3 π₯=log 12πππ3 β4
πππ3 π₯=log 12πππ3 β4
We can now key this into our calculator: log 12 = Γ· log 3 = - 4 =
πππ3 π₯=β1.74Now we must rewrite it.
πβπ.ππ=ππ=.πππ
) = 5Notice we have βlogβ 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
π₯ (π₯+2 )=35
π₯ (π₯+2 )=35Foil, set equal to zero and either use
quadratic formula or factor.
π₯2+2π₯β243=0π₯=
β2Β±β(2)2β4 (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. Donβt forget to FOIL the β x β 2β
After moving the x and -4 over we either factor or use quadratic formula.
π₯=β5Β±β(5)2β4 (8)
2
π₯=β5Β±β(5)2β4 (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
(π₯+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)
Cancelβ¦.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 =
π₯=β13Β±β(13)2β4 (60)
12Since 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)
Cancelβ¦.
(x -4)(x+2) = (x-4)(x+2)
(x -4)(x+2) = (x-4)(x+2)
4(x β 4) = 5 +Distribute, collect like terms, set equal to zero, factor or use quadratic formula.
4x β 16 = 5 +4x β 16 =
0= π₯=
β8Β±β(β8)2β4(10)4
π₯=β8Β±β244
ππ π₯=β4 Β±β62
π¦=π₯3+8 π₯2+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 pβs and qβs 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.
Β± 1, Β±2, Β±3, Β±4, Β±6, Β±12
Step 2: Now using synthetic division we see which factor divides evenly into the polynomial.
π¦=π₯3+8 π₯2+19π₯+12Trial # 1: x = 1
1 1 8 19 12
119
928
28
40Remember that if this
doesnβt result in a 0 then x = 1 IS NOT a factor.
π¦=π₯3+8 π₯2+19π₯+12Trial # 2: x = -1
-1 1 8 19 12
1-17
-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.
π₯2+7 π₯β+12(x + 3)(x + 4)
That means the solutions are: -1,-3, -4