Re: I just need help



fishfry wrote:
> In article <1nLre.90764$1V1.1341546@xxxxxxxxxxxxxxxxxxx>,
> "Stephane Hebert" <stephhebert@yourpants~videotron.ca> wrote:
>
> > Hi folks,
> >
> > I've been doing some math these last days and man am I rusty <g>
> >
> > Now i'm stumped on how to figure this one out:
> >
> > 0.25811 = 10^(x/20)
> >
>
>
> 0.25811 = 10^(x/20)
>
> log_10(.25811) = x/20 [taking base-10 log of both sides]
>
> 20*log_10(.25811) = x [multiply both sides by 20]
>
> [log_10 means "log base 10"]
>
> Now it's calculator time. I entered "scientific calculator" into Google
> and the first link was http://www.calculator.com/calcs/calc_sci.html
>
> I entered .25811 and pressed the "log" button. I got -0.58819516918458.
> Then I pressed "x" and "20" and got -11.7639033836915


Or you could try the zeroth link - try "searching" for
20 * log(0.25811)



>
> Of course the true answer should be written -11.7639033836915... with
> three dots at the end to indicate that the result is only an
> approximation. In fact in math class, 20*log_10(.25811) would be
> regarded as the correct answer.
>
>
>
>
> > I primarily need to isolate x.
> >
> > I guess the above could also be written...
> >
> > 0.25811 = ??10^x <<-- that's 20th root of 10^x, if that's ever possible.
> >
> > ...but i'm still just as stumped as I was a couple of lines above <g>
> >
> > How the hell can I get something like x = ???
> >
> > BTW, I know that in this particular case x = -12.
> >
> > Thanks !
> >
> > Steph

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