- #1

roboredo

- 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

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- Thread starter roboredo
- Start date

- #1

roboredo

- 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

- 13,256

- 1,249

What do you mean by "derivate this equation" ?

Daniel.

Daniel.

- #3

roboredo

- 8

- 0

- #4

- 13,256

- 1,249

2.The slope is given by the derivative

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

Daniel.

- #5

roboredo

- 8

- 0

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

- #6

roboredo

- 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

- 13,256

- 1,249

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

Daniel.

Daniel.

- #8

roboredo

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- #9

- 13,256

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

and now use the derivative of the natural logarithm.

Daniel.

- #10

roboredo

- 8

- 0

i am still struggling!

- #11

- 13,256

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

Daniel.

- #12

roboredo

- 8

- 0

I still haven't figured it out...

but thank you any way!

but thank you any way!

- #13

- 35,995

- 4,714

:)

Zz.

- #14

da_willem

- 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

roboredo

- 8

- 0

Thank You!

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