- #1

- 8

- 0

Can sameone help me with the derivate of this equation?

y=a log10 (x) + B

Thank you

Marta

y=a log10 (x) + B

Thank you

Marta

- Thread starter roboredo
- Start date

- #1

- 8

- 0

Can sameone help me with the derivate of this equation?

y=a log10 (x) + B

Thank you

Marta

y=a log10 (x) + B

Thank you

Marta

- #2

- 13,001

- 550

What do you mean by "derivate this equation" ?

Daniel.

Daniel.

- #3

- 8

- 0

- #4

- 13,001

- 550

2.The slope is given by the derivative

[tex] y(x)=a\lg x+b \Rightarrow \frac{dy(x)}{dx}=...? [/tex]

Daniel.

- #5

- 8

- 0

But.. do you actually know the derivate of this equation?...

- #6

- 8

- 0

I am supposed to fit some data into this model

y= a log10 (x) + B, where a and b are constants

then I should calculate the slope of the isotherm in order to obtain an index

can someone help me?

Thank you , marta

- #7

- 13,001

- 550

What's the connection between the logarithm base 10 and the logarithm base "e" ?

Daniel.

Daniel.

- #8

- 8

- 0

- #9

- 13,001

- 550

[tex] \lg x= \frac{\ln x}{\ln 10} [/tex]

and now use the derivative of the natural logarithm.

Daniel.

- #10

- 8

- 0

i am still struggling!

- #11

- 13,001

- 550

[tex] \frac{d}{dx}\left(a\frac{\ln x}{\ln 10}\right)=\frac{a}{\ln 10}\frac{d \ln x}{dx} [/tex]

Daniel.

- #12

- 8

- 0

I still haven't figured it out...

but thank you any way!

but thank you any way!

- #13

- 35,847

- 4,664

:)

Zz.

- #14

- 599

- 1

The answer is (for x>0)

[tex]\frac{a}{ln10} \frac{1}{x}=(\frac{a}{2.302585...} ) \frac{1}{x}[/tex]

[tex]\frac{a}{ln10} \frac{1}{x}=(\frac{a}{2.302585...} ) \frac{1}{x}[/tex]

Last edited:

- #15

- 8

- 0

Thank You!

- Replies
- 1

- Views
- 922

- Replies
- 1

- Views
- 2K