Question

simplify the expression

Answer 1

#20

Answer 2

#20
a.\[2^{8\log_2(x)}\]

first, move the 8 to become the power of x

Answer 3

then use \[\large b^{\log_b A} =A\]

Answer 4

so \[2^{\log _{2^{}}x ^{8}}\]

Answer 5

good. now cancel the base and the log

Answer 6

how do i do that?

Answer 7

like the formula i gave above ,
except the base as 2,
and the argument of the log is x^8

Answer 8

i dont quite understand...

Answer 9

\[\large b^{\log_b A} =A\\
\large 2^{\log_2 (x^8)} =?\]

Answer 10

just solve for x now?

Answer 11

no, you are simplifying the original expression

Answer 12

\[2^{8\log_2(x)}\\
=2^{\log_2(x^8)}\\=\qquad\qquad(\small\text{some expression in terms of } x)\]

Answer 13

can we go step by step through this im not really understanding.

Answer 14

which step dont you understand, you got the first one right?

Answer 15

simplifying it further from \[2^{\log _{2}x ^{8}}\]

Answer 16

just cancel the base with the log

\[b^{\log_b(A)}\\
=\cancel{b^{\log_b}}\,^{(A)}\\
=A\]

Answer 17

so it just cancels out and the answer is x^8

Answer 18

yes, note that you can only cancel because the base of the exponential is the same and the base of the log

Answer 19

oh ok that makes sense. so as in problem D and E since they are not the same how does that work?

Answer 20

you'll have to use a change of base formula i think

\[\log_bx=\frac{\log_cx}{\log_cb}\]

for D you'll want to choose \(c\) to be 2

Answer 21

what do you mean?

Answer 22

after you move the 3 into the index of x

change \(\log_{1/2} x^3\) into \(\dfrac{\log_2(x^3)}{\log_2(1/2)}\)

Answer 23

what happened to the 2 in front?

Answer 24

that 2 is still there, we are just working on its index at the moment,

Answer 25

now can you simplify the denominator log_2(1/2) ?

Answer 26

(it might help to rewrite 1/2 as some power of 2)

Answer 27

so i got to changing \[\log_{1/2}x^3 \] into the other on as shown above and i dont know how it went to \[\log_{1/2} \to \log_{2} \] for both the numerator and denominator

Answer 28

the Change of Base Formula can be used for any new base \(c\),
choosing \(c\to2\) mean that we will be able to cancel with the 2 out the front later

Answer 29

ok. so how do we simplify this from here \[\log_{2}(x^3)/\log_{2}(1/2) \]

Answer 30

lets look at the denominator ,

\[\log_2(1/2)\]

this can be simplified, if you first express 1/2 as some power of 2

1/2 = 2^?

Answer 31

.5

Answer 32

no nvm lol doesnt work

Answer 33

yes!, so that means

\[\log_2(1/2) = \log_2(2^{0.5}) =\]

can you simplify this further now?

Answer 34

remember \[\log_b A^n = n\log_bA\]

and\[\log_bb =1\]

Answer 35

this is where the 2 infront comes into play huh. it cancels out the bases

Answer 36

not yet, that 2 cancels soon, but not yet

Answer 37

\[.5\log_{2}2 \]

Answer 38

\(\large \frac 12 \ne 2^{0.5}\)

\(\large \frac 12 = \frac {1}{2^{1}} = 2^{-1} \)

Answer 39

whoops

Answer 40

???

Answer 41

do it would be \[-1\log_{2}2 \]

Answer 42

we made a mistake, \(2^{1/2}\) is \(\sqrt 2\)
\(2^{-1}\) is \(1/2\)

Answer 43

that's better, the numerator becomes \( -1\log_2(2)\)

now can you simplify \(\log_2(2)\) ?

Answer 44

.60

Answer 45

?

Answer 46

no lol sorry i tried putting it into the calculator and it didnt really worked out

Answer 47

for any base\[\log_b(b)=1\]this is like \[b=b^1\]

Answer 48

so \[\log_2(2) =\,\dots\]

Answer 49

2^1

Answer 50

2

Answer 51

nope

Answer 52

1?

Answer 53

yes, log_2(2) = 1

Answer 54

ok

Answer 55

So this makes \[-1\log_2(2)\] = ?

Answer 56

-1*1=-1

Answer 57

so we have

\[\log_{1/2}(x^3)=\dfrac{\log_2(x^3)}{\log_2(1/2)}=\dfrac{\log_2(x^3)}{-1}\]

Answer 58

yes no we simplify the top?

Answer 59

we can move the -1 somewhere , where can we put it?

Answer 60

remember that \(\frac1{-1}\)=-1

Answer 61

i have no clue

Answer 62

\[=\dfrac{\log_2(x^3)}{-1}\\
=-\log_2(x^3)\] right?

now used \(n\log_b(A)=\log_b(A^n)\)

Answer 63

yes

Answer 64

\[-3\log_{2}x \]

Answer 65

that is right, but we want to go the other way, we want the -3 as the index of the x

Answer 66

\[\log_{2}x^-3 \]

Answer 67

yep

Answer 68

\[\log_{2}x^{-3} \]

Answer 69

so we have shown that
\[\log_{1/2}(x^3)=\dfrac{\log_2(x^3)}{\log_2(1/2)}=\dfrac{\log_2(x^3)}{-1}=\log_2(x^{-3})\]

Answer 70

so then the 2 comes into play and wala \[x ^{-3}\]

Answer 71

YESS!!! \[\huge \color{red}\checkmark\]

Answer 72

sweet so would we work e the same way?

Answer 73

e)

Answer 74

yeah , E is very similar to D,

Answer 75

so how can i find or express\[\log_{3}1/3 \]

Answer 76

1/3 = 3^?

Answer 77

it would be \[\log_{3}3^{-?} \]

Answer 78

yeah log_3(1/3) = log_3(3^{-1})

Answer 79

why -1?

Answer 80

\[\frac1x = x^{-1}\]

Answer 81

ok so i have \[\log_{3}x ^{-1}/-1 \]

Answer 82

so my answer to e) is \[x ^{1}\]or simply x?

Answer 83

great work

Answer 84

awesome thank you so much. you have been such a great help!

Answer 85

\[\large\color{green} {\ddot\smile}\]

Answer 86

hey on my attachment can you see #22. does th e and the ln cancel each other out and leave me 4x?

Answer 87

Yes!, \[\ln(A) =\log_e(A)\]
so that \[e^{\ln(A)}=A\]

Answer 88

\[e ^{4\ln x}\]

Answer 89

except its not equal to 4x, , you have to move the 4 into the index of the x before you cancel the e with the ln

Answer 90

so it is \[x ^{4}\]

Answer 91

correct

Answer 92

so in be x is squared so would it look like\[e ^{\ln (x ^{2}+1)^{3}}\]

Answer 93

B) not be

Answer 94

yeah that is the right first step for 22. b.

Answer 95

so simplified it would be \[x ^{4}+1\]

Answer 96

sorry x^5

Answer 97

how'd you get that?