Solve for n

Solve for n:

1000= 50 x(1.00125)^n + 10 x(1.00125^n - 1/1.00125 - 1)

Please show your steps.

Let a = 1.00125

$\small{ \begin{array}{rcll} 1000 &=& 50 \cdot (1.00125)^n + 10 \cdot (1.00125^n - \frac{1}{1.00125} - 1) \qquad & | \qquad a = 1.00125\\ 1000 &=& 50 \cdot a^n + 10 \cdot (a^n - \frac{1}{a} - 1) \qquad & | \qquad :10\\ 100 &=& 5 \cdot a^n + (a^n - \frac{1}{a} - 1)\\ 100 &=& 5 \cdot a^n + a^n - \frac{1}{a} - 1\\ 100 &=& 6 \cdot a^n - \frac{1}{a} - 1 \qquad & | \qquad +1\\ 101 &=& 6 \cdot a^n - \frac{1}{a} \\\\ 6 \cdot a^n - \frac{1}{a} &=& 101 \qquad & | \qquad + \frac{1}{a}\\\\ 6 \cdot a^n &=& 101 + \frac{1}{a} \qquad & | \qquad :6\\\\ a^n &=& \dfrac { 101 + \frac{1}{a} }{ 6 } \qquad & | \qquad \log_{10}()\\\\ \log_{10}(a^n) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right)\\\\ n\cdot \log_{10}(a) &=& \log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) \qquad & | \qquad \log_{10}(a)\\\\ n &=& \dfrac{\log_{10} \left( \dfrac { 101 + \frac{1}{a} }{ 6 } \right) }{ \log_{10}(a)} \qquad & | \qquad a = 1.00125\\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + \frac{1}{1.00125} }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101 + 0.99875156055 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} \left( \dfrac { 101.998751561 }{ 6 } \right) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ \log_{10} ( 16.9997919268 ) }{ \log_{10}(1.00125)} \\\\ n &=& \dfrac{ 1.23044360575 }{ 0.00054252909} \\\\ \mathbf{n} & \mathbf{=} & \mathbf{2267.97718911} \end{array} }$

Solve for n over the real numbers:
1000 = 50 1.00125^n+10 (1.00125^n-1.99875)

50 1.00125^n+10 (1.00125^n-1.99875) = 2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801):
1000 = 2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801)

1000 = 2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801) is equivalent to 2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801) = 1000:
2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801) = 1000

2^(1-5 n) 25^(1-n) 801^n+10 ((801/800)^n-1601/801)  =  -16010/801+2^(1-5 n) 5^(1-2 n) 801^n+2^(1-5 n) 25^(1-n) 801^n:
-16010/801+2^(1-5 n) 5^(1-2 n) 801^n+2^(1-5 n) 25^(1-n) 801^n = 1000

2^(1-5 n) 5^(1-2 n) 801^n = e^(log(2^(1-5 n))) e^(log(5^(1-2 n))) e^(log(801^n)) = e^((1-5 n) log(2)) e^((1-2 n) log(5)) e^(n log(801)) = exp((1-5 n) log(2)+(1-2 n) log(5)+n log(801)) and 2^(1-5 n) 25^(1-n) 801^n = e^(log(2^(1-5 n))) e^(log(25^(1-n))) e^(log(801^n)) = e^((1-5 n) log(2)) e^((1-n) log(25)) e^(n log(801)) = exp((1-5 n) log(2)+(1-n) log(25)+n log(801)):
-16010/801+exp(log(2) (1-5 n)+log(5) (1-2 n)+log(801) n)+exp(log(2) (1-5 n)+log(25) (1-n)+log(801) n) = 1000

Simplify and substitute x = exp((1-5 n) log(2)+(1-2 n) log(5)+n log(801)):
 -16010/801+exp((1-5 n) log(2)+(1-2 n) log(5)+n log(801))+exp((1-5 n) log(2)+(1-n) log(25)+n log(801))  =  6 e^((1-5 n) log(2)+(1-2 n) log(5)+n log(801))-16010/801  =  6 x-16010/801  =  1000:
6 x-16010/801 = 1000

Add 16010/801 to both sides:
6 x = 817010/801

Divide both sides by 6:
x = 408505/2403

Substitute back for x = exp((1-5 n) log(2)+(1-2 n) log(5)+n log(801)):
exp(log(2) (1-5 n)+log(5) (1-2 n)+log(801) n) = 408505/2403

Take the natural logarithm of both sides:
log(2) (1-5 n)+log(5) (1-2 n)+log(801) n = log(408505/2403)

Expand and collect in terms of n:
(-5 log(2)-2 log(5)+log(801)) n+log(2)+log(5) = log(408505/2403)

Subtract log(2)+log(5) from both sides:
(log(801)+(-5 log(2)-2 log(5))) n = log(408505/2403)+(-log(2)-log(5))

Divide both sides by -5 log(2)-2 log(5)+log(801):
Answer: |  n = (-log(2)-log(5)+log(408505/2403))/(-5 log(2)-2 log(5)+log(801))=~2268

IT DOES BALANCE, BUT THERE MUST BE A SIMPLER WAY!!. MAYBE CPhill has one.