Derivative of s(t) = (976(.835)^t -1) + 176t using log differentiation

[LearninDaMath]

What is log_b(A) ?

log_b(A) is the exponent that you use with a base of b to get A as a result.

Okay, that is kind of a tricky concept but I get it the idea of the property. I get the fact that A = e^ln(A). However in regard to the function:

s(t) = 976[.835^t - 1] + 176t

I'm trying to figure out how to apply it. The way I'm understanding this property is that it is used when trying to eliminate a log from an equation.

And so I'm not exactly sure how it can be applied to find the derivative.

If I have fx = b^x,

and I take ln of both sides: lnfx = ln(b^x)

Should I apply the property at this point to both sides: e^(lnfx) = e^ln(b^x)

Then I get e^(lnfx) = e^xlnb

And then, I would guess maybe to just cancel out the e's and: lnfx = xlnb.

And then just evaluate the derivative: (lnfx)' = (xlnb)'

i doubt this is correct, it just feels like I'm going in circles or something.

 

[SammyS]

Okay, that is kind of a tricky concept but I get it the idea of the property. I get the fact that A = e^ln(A). However in regard to the function:

s(t) = 976[.835^t - 1] + 176t

I'm trying to figure out how to apply it. The way I'm understanding this property is that it is used when trying to eliminate a log from an equation.

...

No.

It's used so that you have e^t rather than 0.835^t .

Of course you can memorize: the derivative of b^t is ln(b)e^t .

 

[LearninDaMath]

No.

It's used so that you have e^t rather than 0.835^t .

Of course you can memorize: the derivative of b^t is ln(b)e^t .

Okay, so if f(x) = b^x, then its derivative equals f'x = (lnb)(b^x)

So how is it possible that f'(x) could ever equal lnb(e^x)?

In otherwords supposing b = 10, then how can ln10(10^x) be equivalent to ln10(e^x)?

EDIT: I'm sure that ln10(10^x) ≠ ln10(e^x),

So I'm trying to understand what connection I should be making.

 

[LearninDaMath]

Also, would I be raising both sides of the equation to make each side "e to the power of?" since whatever you do to one side of the equation, you do to the other.. and at what point would I apply this? Should I isolate .835^t to one side, then apply e to both sides, then take derivative? Or some other order? Or is isolating .835^t not even the way to start?

 

[SammyS]

Okay, so if f(x) = b^x, then its derivative equals f'x = (lnb)(b^x)

So how is it possible that f'(x) could ever equal lnb(e^x)?

In otherwords supposing b = 10, then how can ln10(10^x) be equivalent to ln10(e^x)?

EDIT: I'm sure that ln10(10^x) ≠ ln10(e^x),

So I'm trying to understand what connection I should be making.

Sorry about that. I had a brain cramp !

Should have been,

$\displaystyle f'(x)=\ln(b)\,e^{x\,\ln(b)}=\ln(b)\,b^x$​