3e^2x-2 ^x+1=0

Answers needed pls

HI ,
THE SOLUTION TO THE PROBLEM:

YOU CAN MULTIPLY BOTH SIDES BY (LN) AND HAVE:

ln(3.e^2x) = ln 2^(x+1)
but ln*e^x = x then, ln*e^2x = 2x.

therefore, 3*2x = (x+1)ln 2
so, 6x = xln2 + ln2
and finally from algebraic manipulations:

x = ln2/(6- ln2)

thanks.

intelligent analysis there hydrochloric acid. Cheers

thanks man. that was a good question there.

@hcl_acid
pls kindly solve this one too. make r the subject of P=a[1-(1+r)^-n]/r

sorry zebra for the delay, i'd post the solution as soon as possible. viewing the question, i could see that you can solve it using the taylor series or laurent. see ya soon,

hcl_acid:

sorry zebra for the delay, i'd post the solution as soon as possible. viewing the question, i could see that you can solve it using the taylor series or laurent. see ya soon,

please hlp me solve it, i need it urgently. tanx.

@hcl_acid

How areu doing? i'm still waiting for the solution to the problem. make r the subject of P=a[1-(1+r)^-n]/r

E be like say this one pass HCl, maybe H2SO4 can solve it!!

Obinoscopy:

E be like say this one pass HCl, maybe H2SO4 can solve it!!

Guy u are funny . But i'm serious i need a solution to the problem, and i wonder where hcl has gone to.

sorry guys for the absence, and as for you zebra the equation you actually posted has no solution since it is composed of Taylor expressions up to the nth degree. it would be useless for me to try to type down these hugh general expressions. therefore i'l recommend that you pick a book on calculus vol. 2 which explains tylor and laurent series. ok?