Please help me with the following sum:
Dermine n if:
665/24=(4/3((3/2)^n-1))/(1/2)
$\displaystyle
\frac{665}{24}
=
\frac
{
\frac{4}{3}.
(\frac{3}{2})^{n-1}
}
{\frac{1}{2}}
$
$\displaystyle
\frac{665}{24}.\frac{1}{2}
=
\frac{4}{3}.
(\frac{3}{2})^{n-1}
$
$\displaystyle
\frac{665}{24}.\frac{1}{2}.\frac{3}{4}
=
(\frac{3}{2})^{n-1}
$
Now take log on both sides and proceed.
Is this the problem?
$\displaystyle \frac{665}{24}=\frac{4}{3}(\frac{(\frac{3}{2})^n-1}{\frac{1}{2}})$
If so, how have you tried to distribute so far? Do you know that dividing by a half is the same as multiplying by 2?
Start by dividing both sides by 4 and multiplying both sides by 3 to give:
$\displaystyle \frac{665}{24}\times\frac{3}{4}=\frac{(\frac{3}{2} )^n-1}{\frac{1}{2}}$
$\displaystyle \frac{665}{24}\times\frac{3}{4}=2\times ((\frac{3}{2})^n-1)$
I would go further, but I don't know whether this is the equation and nor do I know whether you've followed my working thus far. Can you provide any further contribution?
Edit: whoops, was beaten to it.
solving Quacky's equation.
$\displaystyle \frac{665}{24}\times\frac{3}{4}=2\times ((\frac{3}{2})^n-1)$
$\displaystyle
\frac{665}{24}\times\frac{3}{4}\times\frac{1}{2}
=
(\frac{3}{2})^n-1
$
$\displaystyle
\frac{665}{24}\times\frac{3}{4}\times\frac{1}{2}+1
=
(\frac{3}{2})^n
$
solve LHS and take log on both sides
$\displaystyle
log_e(\frac{665}{24}\times\frac{3}{4}\times\frac{1 }{2}+1)
=
n\times log_e(\frac{3}{2})
$
$\displaystyle
\frac{log_e(\frac{665}{24}\times\frac{3}{4}\times\ frac{1}{2}+1)}{log_e(\frac{3}{2})}
=
n
$