Recent content by elctronoob

Thanks very much for your help Samy and the rest of you fine people, much appreciated

Right folks, I've gone away n come back with an improved effort i think... log3(10x-1) - 2 = log3(x2) =log3((10x-1)/x2) = 2 = (10x-1)/x2 = 32 = 9 9x2 -10x + 1 = 0 =>x = 1, x = 1/9 Does this look correct to you guys or am I missing something again?

Thanks jedishrfu and everyone else...I'm going to walk away from it for a while n watch that video... n come back to it later tonight... 15 year old version of me would be lolling his head off at me now...if lolling was a thing back then

i'm just going round in circles here and getting nowhere I'm afraid ##\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))## I understood when i had a common log base, i could divide all by that log base hence ((10 + x) / 1) - 2 = x2 0 = x2 - x - 8 which doesn't give reasonable answers so i...

Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff... I now have it at (log310x - log31) - 2 = log3x2 Then divide out the log3 10x - 1 - 3 = x2 0 = x2 -10x + 3 I've obviously done something wrong again as this gives pretty silly results :(

its written as 2log3x, so I take it to mean 2 * log3X

Hi folks, I'm revisiting logs for the first time in a long time through distance education and I was wondering if someone could have a look over a question I've answered and let me know if I've done it correctly or if I'm way off please Find x if Log3(10x – 1) – 2 = 2log3x I instantly divide...

Thanks for that gneill, much appreciated

Could someone confirm that my thinking on this question is correct for me please? I'm basically just disregarding the Sin200πt and calculating the Vrms and Irms derived from from the 50 value at the start of the expression, i.e 50V. This seems incorrect to me, could someone just check my logic...

did i just miss that Sin200πt is representative of 1 volt? Going by this conversion site, http://www.learningaboutelectronics.com/Articles/Peak-voltage-calculator.php#rms Vrms = 25 / √2 = 17.7V Irms = 17.7 / 40 = 0.4425 amperes Does this look correct? Thanks for your time and help...

Still a bit confused, so can I say Vpeak = 50Sin200πt / 2 = 25Sin200πt Vrms = 25Sin200πt / √2 This still requires a value for t, no?

Thanks for the reply PWiz...any formula I can find for these calculations seem to require a time. The only one I can find without t is Vpeak = Vrms√2...=> Vpeak/√2 = Vrms Vrms / R = Irms I was thinkin maybe I could half the Vac given in the question, but that still involves a t variable...so...

Homework Statement Sorry for what will likely be considered a noob question but I've been stuck on it for a while and its only at the start of my assignment :( Simple circuit Vac = 50Sin200πt, R1 = 40Ω Calculate the value of the RMS current Homework Equations Vt = Vmax sin2πft = Vmax sinωt...