solve equation for n where n is an exponent

**Rosie** · April 7th 2011, 05:30 AM

Please help me with the following sum:

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

**amul28** · April 7th 2011, 05:54 AM

Originally Posted by Rosie

Please help me with the following sum:

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.

**Quacky** · April 7th 2011, 05:54 AM

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.

**Rosie** · April 7th 2011, 06:16 AM

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.

**amul28** · April 7th 2011, 06:32 AM

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 $

**Rosie** · April 7th 2011, 06:40 AM

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.

**amul28** · April 7th 2011, 06:50 AM

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 $n=6$

**Rosie** · April 7th 2011, 06:54 AM

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

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