# Thread: solve equation for n where n is an exponent

1. ## solve equation for n where n is an exponent

Dermine n if:
665/24=(4/3((3/2)^n-1))/(1/2)

2. Originally Posted by Rosie

Dermine n if:
665/24=(4/3((3/2)^n-1))/(1/2)
$
\frac{665}{24}
=
\frac
{
\frac{4}{3}.
(\frac{3}{2})^{n-1}
}
{\frac{1}{2}}
$

$
\frac{665}{24}.\frac{1}{2}
=
\frac{4}{3}.
(\frac{3}{2})^{n-1}
$

$
\frac{665}{24}.\frac{1}{2}.\frac{3}{4}
=
(\frac{3}{2})^{n-1}
$

Now take log on both sides and proceed.

3. Is this the problem?

$\frac{665}{24}=\frac{4}{3}(\frac{(\frac{3}{2})^n-1}{\frac{1}{2}})$

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:

$\frac{665}{24}\times\frac{3}{4}=\frac{(\frac{3}{2} )^n-1}{\frac{1}{2}}$

$\frac{665}{24}\times\frac{3}{4}=2\times ((\frac{3}{2})^n-1)$

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.

4. Quacky you have the right equation and the rest of your questions I do not understand.
Can you please finish the equation because I truely do not get the right answer.

5. solving Quacky's equation.

$\frac{665}{24}\times\frac{3}{4}=2\times ((\frac{3}{2})^n-1)$

$
\frac{665}{24}\times\frac{3}{4}\times\frac{1}{2}
=
(\frac{3}{2})^n-1
$

$
\frac{665}{24}\times\frac{3}{4}\times\frac{1}{2}+1
=
(\frac{3}{2})^n
$

solve LHS and take log on both sides

$
log_e(\frac{665}{24}\times\frac{3}{4}\times\frac{1 }{2}+1)
=
n\times log_e(\frac{3}{2})
$

$
\frac{log_e(\frac{665}{24}\times\frac{3}{4}\times\ frac{1}{2}+1)}{log_e(\frac{3}{2})}
=
n
$

6. amul I do not know what log is, I have never used it before. I am just matric. The sum falls under series and sequences. So therefor I do not understand what you just did.

7. Originally Posted by Rosie
amul I do not know what log is, I have never used it before. I am just matric. The sum falls under series and sequences. So therefor I do not understand what you just did.
ok
$\frac{665}{24}\times\frac{3}{4}\times\frac{1}{2}+1
=
(\frac{3}{2})^n$

when you solve LHS it solves to $\frac{729}{64}$

$
(\frac{3}{2})^6=(\frac{3}{2})^n
$

since bases are equal powers are equal

so $n=6$

8. thank you very much, I see where I made my mistake.