Using logs to find n (1 Viewer)

liquid_soda · May 3, 2015

I'm having a bit of trouble here
500 000 x 1.004375^n -[10 000 (1.004375^n -1)]/ 0.004375
How do I isolate and find n?

Shadowdude · May 3, 2015

liquid_soda said:
I'm having a bit of trouble here
500 000 x 1.004375^n -[10 000 (1.004375^n -1)]/ 0.004375
How do I isolate and find n?

$500000\times 1.004375^{n} - \frac{\left [ 10000\left ( 1.004375^{n} - 1 \right ) \right ]}{0.04375}$

???

If so, well... you can't because there's no equation to solve.

InteGrand · May 3, 2015

Shadowdude said:
$500000\times 1.004375^{n} - \frac{\left [ 10000\left ( 1.004375^{n} - 1 \right ) \right ]}{0.04375}$

???

If so, well... you can't because there's no equation to solve.

I'm guessing the minus sign was meant to be an equals sign (typo)? (Looks like a consumer arithmetic maths Q.)

liquid_soda · May 3, 2015

Apologies for the typo
$500000\times 1.004375^{n} - \frac{\left [ 10000\left ( 1.004375^{n} - 1 \right ) \right ]}{0.004375} =0$
Yup it's consumer, asking about when the amount owning is equal to zero

Squar3root · May 3, 2015

I got $n = 56.54$ (2dp)

liquid_soda · May 3, 2015

Squar3root said:
I got $n = 56.54$ (2dp)

that's it, ~57 months, how is your working out?

Squar3root · May 3, 2015

liquid_soda said:
that's it, ~57 months, how is your working out?

it was really just algebra

I divided both sides by 10000
Multiplied both sides by 0.004375
Divided every term by 1.004375^n
and got 1.004375^n = 1.28
and used logs to get n = 56.24

liquid_soda · May 3, 2015

thanks man

liquid_soda · May 3, 2015

$500000\times 1.004375^{n} - \frac{\left [ 10000\left ( 1.004375^{n} - 1 \right ) \right ]}{0.004375} =0$
$50\times 1.004375^{n} - ( 1.004375^{n} - 1) =0$
To clarify, am I allowed to take out the bracket then later take $1.004375^{n}$ when factoring?
So that becomes $50 \times 1.004375^{n} - 1.004375^{n} +1 =0$
$1.004375^{n} (50+ 1) +1 =0$
??

Squar3root · May 3, 2015

liquid_soda said:
$500000\times 1.004375^{n} - \frac{\left [ 10000\left ( 1.004375^{n} - 1 \right ) \right ]}{0.004375} =0$
$50\times 1.004375^{n} - ( 1.004375^{n} - 1) =0$
To clarify, am I allowed to take out the bracket then later take $1.004375^{n}$ when factoring?
So that becomes $50 \times 1.004375^{n} - 1.004375^{n} +1 =0$
$1.004375^{n} (50+ 1) +1 =0$
??

where did the "0.004375" term go?

no

liquid_soda · May 3, 2015

Squar3root said:
where did the "0.004375" term go?

no

times it to both sides so $50(0.004375)\times 1.004375^{n} - ( 1.004375^{n} - 1) =0$

Squar3root · May 3, 2015

liquid_soda said:
times it to both sides so $50(0.004375)\times 1.004375^{n} - ( 1.004375^{n} - 1) =0$

if you did the algebra correctly, yes you should arrive at the same answer

liquid_soda · May 3, 2015

sigh, I don't know what to do with the brackets
help...

InteGrand · May 3, 2015

liquid_soda said:
sigh, I don't know what to do with the brackets
help...

Maybe to make it clearer, let's just write $x=(1.004375)^n,a=50(0.004375), R = 1.004375\text{ (so }x=R^n)$ .

Then the equation you got is $ax-(x-1)=0$ .

Therefore, $ax-x+1=0$

$\Rightarrow x(a-1)=-1$

$\Rightarrow x=\frac{1}{1-a}$

$\Rightarrow R^n=\frac{1}{1-a}$

$\Rightarrow \ln (R^n)=\ln\left( \frac{1}{1-a}\right)$

$\Rightarrow n\ln R=-\ln(1-a)$ (using the logarithm law $\ln \left(\frac{1}{t}\right) = -\ln t$ on the RHS)

$\Rightarrow n=- \frac{\ln(1-a)}{\ln R}$

$\Rightarrow n=- \frac{\ln(1-50(0.004375))}{\ln (1.004375)}\approx 56.55\text{ } (\text{2 decimal places})$ .

The integer answer is then $\lceil 56.55\rceil=57$ .

liquid_soda · May 3, 2015

haha, I confused myself big time but got there in the end. Thanks everyone for the help

Using logs to find n (1 Viewer)

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