Log Problems (1 Viewer)

[Avenger6]

Hi, im struggleing with these questions:

Im completly lost as to how you would do question 7. In question 14 however, I used the quotient rule to derive y=e^x/2 and got e^x/4 however the book lists the derivative as e^x/2 again??? Any help is grealty appreciated.

 

[vds700]

Hi, im struggleing with these questions:

Im completly lost as to how you would do question 7. In question 14 however, I used the quotient rule to derive y=e^x/2 and got e^x/4 however the book lists the derivative as e^x/2 again??? Any help is grealty appreciated.

the question isn't showing up for me, u sure u attached it correctly?

To differentiate e^x/2
d/dx(e^x/2) = (1/2) .d/dx(e^x) = e^x/2. You dont need to use the quotient rule when u have e^2 multipied by a constant (in this case, 1/2). Only use the quptient rule if u have x in the denominator, (e.g (e^x)/x).

 

[Avenger6]

The image should be showing...try visiting http://img217.imageshack.us/img217/8628/91735408hu2.jpg. So basically your saying that if it were say (e^x)/5 the derivative would still be (e^x)/5 becuase x is not in the denominator?

 

[vds700]

The image should be showing...try visiting http://img217.imageshack.us/img217/8628/91735408hu2.jpg. So basically your saying that if it were say (e^x)/5 the derivative would still be (e^x)/5 becuase x is not in the denominator?

yep spot on...

hmmm that link doesn't work for me, i just get an error message. Maybe my computer is retarded, hopefully someone else can help u out with the other q's if they can access the q's.

 

[minijumbuk]

log<sub>9[/sup] = log316 / log39

Therefore:
= log32 + log316/log39
= log32 + log34²/log33²
= log32 + 2log32/2log33
= log32 + 2log32/2
= log32 + log32
= log3(2x2)
= log34

Oh yea, and the differentiation of e^x is e^x
You should've learnt that in class.

So for your question, y' = e^x/2
So when m at x=0 is:
m= e⁰/2
m= 1/2

Therefore equation of the tangent is:
y - y1 = m(x-x1)
y - 1/2 = 1/2 (x-0)
y= x/2 + 1/2
x-2y+1=0</sub>

 

[Avenger6]

hmm...im still lost in the first question. First of all, how did you turn log₃2 + log₃4²/log₃3² into log₃2 + 2log₃2/2log₃3...shouldn't it be log₃2 + 2log₃4/2log₃3? Finnaly i don't see how you turned log₃2 + 2log₃2/2log₃3 into log₃2 + 2log₃2/2??? Thanks for the help.

 

[minijumbuk]

Sorry, careless mistake xD
And log_aa = 1 for any number a>0

I'll redo it:

= log₃2 + log₃2⁴/log₃3²
= log₃2 + 4log₃2 / 2log₃3
= log₃2 + 4log₃2/ 2
= log₃2 + 2log₃2
= 3log₃2

 

[Avenger6]

Ok its making much more sense now . Im still stuck on how you made log₃2 + 4log₃2 / 2log₃3 = log₃2 + 4log₃2/ 2...i just dont see how you can make 2 the denominator as 2log₃3 can't be divided into 4log₃2??? Sorry if ive confused you.

 

[lyounamu]

Ok its making much more sense now . Im still stuck on how you made log₃2 + 4log₃2 / 2log₃3 = log₃2 + 4log₃2/ 2...i just dont see how you can make 2 the denominator as 2log₃3 can't be divided into 4log₃2??? Sorry if ive confused you.

He jumped a step. That's why it confused you. Ok, I will explain.

log3(2) + 4log3(2)/2log3(3) = log3(2) + 4log3(2)/2 because log3(3) = 1
= log3(2) + 2log3(2)
= 3log3(2)
Hope that explained it better.

 

[Avenger6]

Much better...thanks alot!

 

[tacogym27101990]

and you can still use the quotient rule for the 2nd question
its just unnessecary
but u should still get the correct derivative

 

[Forbidden.]

You have seen statements like:

"If b is a positive real number and r = p/q, where p is an
integer and q is a non-negative integer, then b^r is the
unique qth root of b^p."

and

"If b > 0 and r is rational with a = b^r, then r = log_ba."

These definitions do NOT allow you calculate 2^√3 or 4^π or worse log₁₀2.


"anything to the power zero is 1"
and
"zero to any power is 0"

so what is 0⁰ ?
The symbol 0⁰ is not defined, but we can evaluate the limit of x^x as x approaches 0 from the right.

Similarly, 1^∞ is not defined, but we can evaluate the limit (1 + x^-1)^x as x approaches infinity.