Question

Problem attached inside :) Drawing tool won't work, so I wrote it and attached it :)

LOGS!!! :/

Please explain how to do this :) I don't really get the steps :/ THank you!! :)

Answer 1

Problem!! :)

Answer 2

For the first one, take the natural log of both sides.

Answer 3

ln7^(x=2)=lne^(17x)
(x+2)ln7 = 17xlne
But remember: lne = 1

Answer 4

sorry, my computer crashed just now haha...

so the total becomes 17x1 ? or does it just stay 17xlne?

Answer 5

17 x 1

Answer 6

ohh okay i see
so it simplifies down to 17?

Answer 7

yes

Answer 8

okay :)

so for the second one, how do you start it off? :/

Answer 9

same way

Answer 10

okay is this it?

(x+3)ln10=...... did I start it off right? :/

Answer 11

On the right side you have ln5e^(7-x) which is ln5 + lne^(7-x)

Answer 12

Yes. That is right for the left side.

Answer 13

oh okay so it goes to this now?

(x+3)ln10=ln5+lne^(7-x)

??

Answer 14

yes

Answer 15

okay, i'm not really sure what to do next though :(

Answer 16

On the right you should put: ln5 +(7-x)lne which is:
ln5 +7-x

Answer 17

Now pick up you calculator and type in ln10 and ln5 and substitute those values and solve the equation.

Answer 18

ermm is it

6.91x=8.61-x

?? not sure if i'm plugging it in right :/

Answer 19

ln5=1.61
ln10=2.30

Answer 20

oh okay so i'm plugging into this right?

(x+3)ln10=ln5+7-x

?? or did i do something wrong?

Answer 21

(x+3)(2.30)=1.61+7-x

Answer 22

ohh okay, so

2.30x+6.9=8.61-x ?

Answer 23

2.30x+6.90=8.61-x
3.30x=1.71
x=.52

Answer 24

ohh okay i see :)

so basically it all goes down to isolating the variable?

Answer 25

so for the last one, am I starting it correctly?

ln4+lne^(2x-3) - 5 = e ?

Answer 26

No. The problem is that you cannot take the ln of only some of the terms.

Answer 27

ohh :/ darn.. then i'm not really sure how to start the last one :(

Answer 28

so ln has to be taken out of all of them?

Answer 29

Add 5 to both sides

Answer 30

so ln4+lne^(2x-3)=e+5 ?

Answer 31

Divide both sides by 4

Answer 32

So now you have:
e^(2x-3)=(e+5)/4

Answer 33

Now take the ln of both sides.

Answer 34

Ohhh so when you divide the 4, the ln goes away?

Answer 35

lne^(2x-3)=ln(e+5)/4 ?

Answer 36

i don't know if i'm doing this right... but i got -23 for the left side... :/ is that right?

Answer 37

Let's start again:
1. Add 5 to both sides
2. Divide both sides by 4
3. Take the ln of both sides
4. Add 3 to both sides
5. Divide both sides by 2

Answer 38

that's based off the original equation?

if so, do we get this?

4e^2x-3=e+5 ?

Answer 39

yes. Now divide both sides by 4

Answer 40

okay so we get e^2x-3=(e=5)/4 ?

Answer 41

e(2x-3)=(e+5)/4

Answer 42

oh it's not to the power of 2x-3 anymore?

Answer 43

Yes it is. I forgot to type that.

Answer 44

ohh okay haha :p so now you said to take the ln?

so lne^(2x-3) = ln(e+5)/4 ?

Answer 45

e^(2x-3)=(e+5)?4

Answer 46

yes

Answer 47

how would i plug this into the calculator?

Answer 48

On the left you have 2x-3 since lne=1

Answer 49

oh yeah :)

so once that happens, it just becomes 2x-3=ln4e+ln4(5) ?

Answer 50

So you have 2x-3=ln(e+5)/4

Answer 51

not sure if that right side is correct haha :/

Answer 52

Now, we know that e is approximately 2.718 so make that substitution and solve

Answer 53

okay

2x-3=ln(2.718+5)/4

wait, so is the ln gone?

Answer 54

you mean on the left?

Answer 55

no on the right hand side? did I put the ln there correctly? or should it just be 2x-3=(2.718+5)/4 ?

Answer 56

No. You haven't found the natural log of the right so you should add 2.718 and 5. Then you should divide that by 4. Then hit the ln button

Answer 57

2x-3=ln(1.9192)
2x-3=.6573

Answer 58

ohh okay so 2x=3.6573

x=1.82865=1.83 ?

Answer 59

yep

Answer 60

okay awesome!! :) i'll have to review all this haha it's confusing!!

thanks so much!! :)