Solving for $n$ in $\frac{1.1^n - 1}{(0.1)1.1^n} =4.6$

Question

I want to solve for $n$ in this equation: $$PWF = \frac{(1+i)^n-1}{i(1+i)^n} = \frac{1.1^n - 1}{0.1 \cdot 1.1^n} =4.6$$

I know that log must be taken of both the right as well as the left sides of the equation but I am confused as to how.

Answer 1

So, what we have is

$$\frac{1.1^n - 1}{0.1 \cdot 1.1^n} =4.6$$

We multiply both sides by $0.1$ and split up the fraction:

$$\frac{1.1^n - 1}{1.1^n} = 1- \frac{1}{1.1^n} = 0.46$$

Solving for the fraction, we get

$$0.54 = \frac{1}{1.1^n}$$

Taking the reciprocal,

$$1.1^n = \frac{1}{0.54} = \frac{50}{27}$$

We take the logarithm, base $1.1$, of both sides.

$$\log_{1.1}(1.1^n) = n = \log_{1.1} \left( \frac{50}{27} \right)$$

From there you pretty much have to plug it into a calculator to get a value. WolframAlpha gives about $6.46506$.

Answer 2

Guide:

Try not to think of you have to apply logarithm from the start. Focus on the simplication and only take logarithm when the form is right.

Let $1.1^n=x$, now multiply both sides by the denominator on the right and we obtain a linear equation. Solve for $x$.

After solving for $x$, the problem becomes $1.1^n=x$ and we have $n=\frac{\ln x}{\ln 1.1}$.