Thread: Logarythimic Proof Help

The question is to prove the "k" in the following queston is equal to: k = 1/10Log_e(10/7)
and the original question is 70 = 100e^-k10

I'll show the steps i have used so far, but i don't understand the next step needed..

> 70/100 = e^-k10
> 7/10 = e^-k10
> log_e(7/10) = -k10

That is as far as i can get, if you know the next step please respond with the Log law that goes with it, Any help would be appreciated.

-Medusa

After further work, i now am just questioning whether a step is mathmatically allowed? Is it possible to go from my last step to

>loge((10/7)^-1) = -k10

and from that just perform basic arithmetic to solve?

>-ln(10/7) = -k10
>1/10ln(10/7) = k

Or is one of the steps incorrect?

All those steps are correct.
You alternatively could have done this: 7/10=e^-k10 So 7/10=1/e^k10 So e^k10=10/7
So k10=ln(10/7) So k= 1/10ln(10/7)