Recent content by blackfriars

Thanks for that all i can do is learn Cheers

Yes i am

hi to all looking for some help to rearrange a formula to make x the subject 2x/5-3b/4=x+5 i moved +5 to the left and it became -5 i thought the negative 5 would cancel the denominator 5 this is what i thought i could do 2x-3b/4=x -2x -2x -3b/4 =-x multiply both sides by -1 and...

Could you show steps for solving for x Thanks - - - Updated - - - Yeah i got it now cheers mate

the answer i got for X was 5 just by using algebra yes it is the latter equation thanks

hi could anyone show me the steps for solving this equation i thought it was to be solved by logs find X if 2^3x+1=32

thank you i could not see that brilliant

hi , in your 3rd line of work where did the 2nd (e^2) come from thanks

hi sorry for the questions but i cannot transpose the formula to make x the subject could you show the workings for making x the subject thanks

hi , hope someone can help as i can't get past a certain step the natural logs is the problem ln⁡((x-1)/(x-3))=2 i can get to this point here -1 = e_x^2-x-3 -1+3=x(ⅇ^2-1) 2 = x(ⅇ^2-1) 2/((ⅇ^2-1) )=x((ⅇ^2-1)/(ⅇ^2-1)) X = 2/(ⅇ^2-1) the solution i got was this x= (2/(ⅇ^2-1)) → 0.3130352855 but the...

sorry this is what i meant , i wrote the equation the wrong way this is the correct way hi , hope someone can help as i can't get past a certain step the natural logs is the problem ln⁡((x-1)/(x-3))=2 i can get to this point here -1 = e_x^2-x-3 -1+3=x(ⅇ^2-1) 2 = x(ⅇ^2-1)...

hi , hope someone can help as i can't get past a certain step the natural logs is the problem ln(x-1/x-3)=2 i can get to this point here -1=x(e^2-1)-3 but the lecturer gave a solution of 3e^2-1/e^2-1 = 3.313035285 how do i get to this the solution i got was this 2/e^2-1 =0.3130352855

hi, yes this is what i initially did and when entering the data into my casio i kept getting an error message this led me to believe there was no solution, but then again i was not 100% sure thanks so much for your help

hi mark , really sorry , but i made the mistake the original equation was 5x+ sqroot2x-3=4 i did not cop the missing addition sign sorry about that

hi Markf, well i have just used the Newton method, it was long, but the result i got was similar to yours this is what i did i inputed 1.6216 as the x value in this eq 50x^3-75x^2-16=15.98... this is fine but when i input the same value into the original eq i get something different...