a company pays a car for 50000 and pays of 7876 in the first month, every other month they pay of 20% more than the month before. How many m

 

$$\\7876+1.2*7876+1.2^2*7876+.......<50000\\
7876+1.2*7876+1.2^2*7876+.......<50000\\
7876(1+1.2+1.2^2+.....................) <50000\\
$In the brackets is the sum of a GP$\\\\
7876(\frac{1.2^n-1}{0.2})>50000\\\\
\frac{1.2^n-1}{0.2}>50000/7876\\\\
1.2^n-1>(50000/7876)*0.2\\\\
1.2^n>\{[(50000/7876)*0.2]+1\}\\\\
log(1.2^n)>log\{[(50000/7876)*0.2]+1\}\\\\
nlog(1.2)>log\{[(50000/7876)*0.2]+1\}\\\\
n>\frac{log\{[(50000/7876)*0.2]+1\}}{log(1.2)}\\\\$$

 

 

$${\frac{{log}_{10}\left(\left(\left({\frac{{\mathtt{50\,000}}}{{\mathtt{7\,876}}}}\right){\mathtt{\,\times\,}}{\mathtt{0.2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{log}_{10}\left({\mathtt{1.2}}\right)}} = {\mathtt{4.495\: \!567\: \!527\: \!544\: \!848\: \!3}}$$

 

So it will take 5 months

It would have been easier to do it without the formulas LOL.

I always do everything the long way.

 

----------------------------------------------------------------------------------

Here is an easier method

 

	The months payment	Total paid
month 1	7876	7876
month 2	7876*1.2=9451.20	7876+9451.2=17327.20
month3	11341.44	28668.64
month 4	13609.73	42278.36
month 5	16331.68	58610.04    TO MUCH