Question

solve for x log7 10-1/3log7 x+log7 2

Answer 1

\[x \log_{7} 10 -\frac{ 1 }{ 3 }\log_{7} x + \log_{7} 2\] Is this your problem?

Answer 2

my bad the x is not there in the front?

Answer 3

also if you want to solve for x it should be = to something. What is missing there?

Answer 4

\[\log_{7} 10 - \frac{ 1 }{ 3 }\log_{7} x + \log_{7} 2 = ??\]

Answer 5

yes

Answer 6

sorry its equal to the log of 7 2

Answer 7

OK so your problem is\[\log_{7} 10 - \frac{ 1 }{ 3 }\log_{7} x = \log_{7} 2\]

Answer 8

do you know the logarithmic rules?

Answer 9

yes thank you so much can u help me and explain?

Answer 10

no
my teacher doesnt teach that much

Answer 11

i have to use the book and im struggling

Answer 12

\[\log_{a} A + \log_{a} B = \log_{a} (A . B)\]

Answer 13

\[\log_{a} A - \log_{a} B = \log_{a} \frac{ A }{ B }\]

Answer 14

umm can you break it down into words too?

Answer 15

Log of A to the base a + Log of B to the base a = Log (A times B) to the base a

Note: The base should be the same for you to do that

Similarly for Log A to the base a - Log B to the base a = Log ( A Divided by B) to the base a.

eg Log 2 + Log 3 = Log ( 2x 3) = Log 6, they both should have the same base. For convenience I did not write the base.

Answer 16

Log 6 - Log 3 = Log (6/3) = Log 2

Answer 17

ok so i would use the first equation

Answer 18

there is 3rd rule

Answer 19

\[m \log_{a} A = \log_{a} A ^{m}\]

Answer 20

eg. 3 Log 2 = log 2³ = log 8

Answer 21

pl. write down these rules somewhere all in one place so you can refer to it whenever you need them

Answer 22

ok thank you soo soo mcuh one last thing could you solve an equation like mine so i have an example

Answer 23

4th rule\[\log_{a} a = 1 ; ( example.) \space \log_{5} 5 = 1\]

Answer 24

fifth rule

Answer 25

Log 1 to any base = 0

Answer 26

All these five rules write them down

Now we will use these rules to solve your problem

Answer 27

ok i wrote them

Answer 28

now in your problem ( in any problem ) first thing you check is , Is all the bases the same. Here in your problem all logs on either side have the same base which is 7.

So from now on for this problem if I do not write the base it is understood that the base is 7 ok.

Answer 29

ok

Answer 30

Now we have 3 logs in the problem .

The first one Log 10 we cannot simplify any further. So we keep it

The second one 1/3 log x. This is of the form m log A, m = 1/3 and A = x.

Now tell me which rule do you use to rewrite it?

Answer 31

3rd rule

Answer 32

yes . can you rewrite it?

Answer 33

ill try just a sec

Answer 34

\[-1/3\log _{7}x=\log _{7}x ^{-1/3}\]

Answer 35

?

Answer 36

in a way yes. but I will leave the negative sign alone and just keep a positive 1/3 for the power.

Answer 37

\[\log_{7} 10 - \frac{ 1 }{ 3 }\log_{7} x = \log_{7} 10 - \log_{7} x ^{\frac{ 1 }{ 3 }}\]

Answer 38

ok so subtraction becomes divison?

Answer 39

absolutely

Answer 40

remember we are just on our left hand side of the equation, we have not gone to the right yet

Answer 41

so let us go ahead and further simplify these 2 logs into one log using rule no. 2

Answer 42

ooh ok

Answer 43

can you do it?

Answer 44

im gonna try

Answer 45

good I 'll wait

Answer 46

\[\log _{7}10-\log _{7}x=\log _{7}\frac{ 1 }{ 3 }\]

Answer 47

o.o

Answer 48

you can just type it out for now

Answer 49

ok

Answer 50

you try the tools for this later

Answer 51

k

Answer 52

cool thanks for the medal does that mean i got that part right sorta?

Answer 53

no... I just like the way you try things out. but that was way wrong

Answer 54

oooh can u show me how then?

Answer 55

\[\log_{7} 10 - \log_{7} x ^{\frac{ 1 }{ 3 }} = \log_{7} \frac{ 10 }{ x ^{\frac{ 1 }{3 }} }\]

Answer 56

Log A - Log B = Log (A/B), A = 10 and B = x^1/3

Answer 57

now that is your Left hand side of the equation. remember your right hand side was log 2 to the base 7?

Answer 58

ya

Answer 59

Ok now since both sides have one log and both are same base. We can drop the log like this.

if Log A = Log B then A = B

Answer 60

what?how are they equal

Answer 61

That is the rule no. 6

I\[\log_{a} A = \log_{a} B ; then \space A = B\]

Answer 62

ok i understand that but how are they equal

Answer 63

we still have to solve for x

Answer 64

ok

Answer 65

\[\log_{7} (\frac{ 10 }{ x ^{\frac{ 1 }{ 3 }} }) = \log_{7} 2\]means\[\frac{ 10 }{ x ^{\frac{ 1 }{ 3 }} } = 2\]

Answer 66

Now cross multiply

Answer 67

10/2 = x^1/3

Answer 68

5 = x^1/3

Answer 69

\[x^{\frac{ 1 }{ 3 }} = 5\]

Answer 70

125?

Answer 71

now cube both sides

Answer 72

yes

Answer 73

yay now i feel sorta smart

Answer 74

x = 125

Answer 75

So use the rules I gave you and you will always use a couple of them and solve your x

Answer 76

well done

Answer 77

ok

Answer 78

\[\left[\begin{matrix}x+y= & 9\\ 4x-5y= & 0\end{matrix}\right]\]

Answer 79

hw do u inverse this matrixs

Answer 80

could you pl. post it in a new window?