Question

e^(2x + 4) - 7 = 0
solving for x
log( e^(2x + 4))=log(7)
2x+4=log(7)
x=(log(7)-4) /2

Answer 1

is this right or not?

Answer 2

it is right

Answer 3

cause i tried this answer in my quiz and its wrong

Answer 4

but it also says in the question "Solve each equation for x. Write the exact value: do not write an approximation."

Answer 5

does that mean just a value?

Answer 6

try \(\large \frac{1}{2}(\ln 7 - 4)\)

Answer 7

the online grader is much dumber than you think it is.. you need to try all possible variations sometimes...

Answer 8

i only get 3 tries and i used up 1

Answer 9

then try try this : \(\large \frac{1}{2}(\ln( 7) - 4)\)

Answer 10

it also says "Note: To see what your typed text represents, click on the picture of an eye to the right of the answer box. However, if you use the allowed exp(x) for e^x, the eye will still display it as exp(x) rather than as e^x.

Answer 11

does that mean i have to represent the answer in exponential form?

Answer 12

i don't think its that stupid.. pray ur best gods and try above form >.<

Answer 13

but wait a sec, why are you doing this algebra problem in calculus course ?

Answer 14

lol i'll just ask my lecturer 2moro, can't afford to get it wrong

Answer 15

i think they r just trying to get us to learn the log rules before we do integration with logs

Answer 16

also how would i do the above question?

Answer 17

differentiate implicitly both sides

Answer 18

no the first one isn't asking for derivative

Answer 19

\[\large e^{x^7y} = x+y\]

Answer 20

differentiate both sides with respect to x, implicitly

Answer 21

\[\large \left( e^{x^7y} \right)'= \left(x+y\right)'\]

Answer 22

\[\large e^{x^7y} \left( x^7y \right)'= 1 + y'\]

Answer 23

can you take it home from here ?

Answer 24

i have forgot how to do implicit differentiation ,let me look back at my notes

Answer 25

you just differentiate as you would differentiate product of two different functions of x : f(x)*g(x)

Answer 26

y = g(x) here, so u need to apply chain rule

Answer 27

is the first step clear ?

Answer 28

derivative of e^x is just e^x

so derivative of \(\large e^{x^7y} \) is just \(\large e^{x^7y} \)
not exactly, you need to use chain rule because the exponent is not just x

Answer 29

\(\large \left(e^{x^7y} \right)' = e^{x^7y} \left(x^7y \right)' \)

Answer 30

if that makes any sense..

Answer 31

but how do u (x^7y)'

Answer 32

we use product rule

Answer 33

\[\large \left(fg\right)' = f'g + fg'\]

Answer 34

\[\large \left(x^7y\right)' = \left(x^7\right)' y + x^7 \left(y\right)'\]

Answer 35

7x^6y _x^7(y)'

Answer 36

from where did _ came from hmm

Answer 37

\[\large \left(x^7y\right)' = \left(x^7\right)' y + x^7 \left(y\right)' = 7x^6y + x^7y'\]

right ?

Answer 38

\[\large e^{x^7y} \left( x^7y \right)'= 1 + y'\]

\[\large e^{x^7y} \left(7x^6y+x^7y'\right)= 1 + y'\]

Answer 39

isolate \(\large y'\)

Answer 40

how did u get 1+y'

Answer 41

thats the derivative of right hand side : x + y

Answer 42

sorry i was trying to find my notes on this

Answer 43

i tried to isolate y', but failed how would u of it?

Answer 44

@Kainui

Answer 45

\(\large e^{x^7y} \left(7x^6y+x^7y'\right)= 1 + y' \)


\(\large e^{x^7y} \cdot 7x^6y+ e^{x^7y} \cdot x^7y' = 1 + y' \)


\(\large e^{x^7y} \cdot 7x^6y - 1 = y'\left(1 - e^{x^7y} \cdot x^7 \right) \)


\(\large \dfrac{e^{x^7y} \cdot 7x^6y - 1 }{1 - e^{x^7y}\cdot x^7} = y'\)

Answer 46

something like that maybe ? if grader cries, you may substitute \(\large e^{x^7y}\) by \(\large x+y\) so that :

\(\large \dfrac{(x+y) \cdot 7x^6y - 1 }{1 - (x+y)\cdot x^7} = y'\)

Answer 47

why is e^x^7y substituted with (x+y)?