How can I evaluate this exponential equation with natural logarithm $6161.859 = 22000\cdot(1.025^n-1)$?

Question

I'm trying to evaluate an exponential equation with natural logarithm, but I'm certainly doing something wrong, can someone explain me how would you solve it using natural logarithm?

$$6161.859 = 22000\cdot (1.025^n-1)$$

Expected result: $\approx 10$

Thanks in advance.

Answer 1

First we need to isolate the exponential term:
$$0.2800845 = 1.025^n - 1$$ $$1.2800845 = 1.025^n$$ Now we can take logarithms. I'll use the natural logarithm since it's more widely used, but you can use any one you like (such as $\log_{1.025}{(x)})$
Now we have: $$\ln{(1.2800845)} = \ln{(1.025^n)}$$ From here we can use the fact that $\ln{(a^n)} = n\ln{(a)}$ to remove $n$ from the logarithm. Divide by the logarithm on $n$'s side, and compute.

For the record, we get numerically that $n \approx 9.999998602$.

Answer 2

we get by dividing $\frac{6161.859}{22000}=1.025^n-1$
adding $1$ on both sides we get $\frac{6161.859}{22000}+1=1.025^n$
taking the logarithm of both sides we obtain $\ln\left(\frac{6161.859}{22000}+1\right)=n\ln(1.025)$
from here you will get $n$
with the given precision we obtain $n\approx 9.999998602$