How can I evaluate this logarithmic expression?

Question

I am solving a problem about recursion trees and I applied the geometric series sum formula and I have the following expression: $$ \left(\frac{6}{7}\right)^{\log_{\frac{7}{5}}(N) + 1} $$

How can I write this in a simpler form in terms of N?

Answer 1

Well, if I understand right you have:

$$x:=\left(\frac{6}{7}\right)^{\log_{\frac{7}{5}}\left(\text{n}\right)+1}\tag1$$

Using the rule:

$$\text{a}^{\text{b}+\text{c}}=\text{a}^\text{b}\cdot\text{a}^\text{c}\tag2$$

We can rewrite equation $\left(1\right)$ as follows:

$$x=\left(\frac{6}{7}\right)^1\cdot\left(\frac{6}{7}\right)^{\log_{\frac{7}{5}}\left(\text{n}\right)}=\frac{6}{7}\cdot\left(\frac{6}{7}\right)^{\log_{\frac{7}{5}}\left(\text{n}\right)}\tag3$$

Using the rule:

$$\log_\text{a}\left(\text{b}\right)=\frac{\ln\left(\text{b}\right)}{\ln\left(\text{a}\right)}\tag4$$

We can rewrite equation $\left(3\right)$ as follows:

$$x=\frac{6}{7}\cdot\left(\frac{6}{7}\right)^\frac{\ln\left(\text{n}\right)}{\ln\left(\frac{7}{5}\right)}\tag5$$

Using the following rules:

• $$\text{a}^\frac{\text{b}}{\text{c}}=\left(\text{a}^\frac{1}{\text{c}}\right)^\text{b}\tag6$$
• $$\ln\left(\frac{\text{a}}{\text{b}}\right)=\ln\left(\text{a}\right)-\ln\left(\text{b}\right)\tag7$$

We can rewrite equation $\left(5\right)$ as follows:

$$x=\frac{6}{7}\cdot\left(\left(\frac{6}{7}\right)^\frac{1}{\ln\left(7\right)-\ln\left(5\right)}\right)^{\ln\left(\text{n}\right)}\tag8$$

Now, using an approximation we can write:

• $$\frac{6}{7}\approx0.85714\tag9$$
• $$\left(\frac{6}{7}\right)^\frac{1}{\ln\left(7\right)-\ln\left(5\right)}\approx0.63246\tag{10}$$

So:

$$x\approx0.85714\cdot0.63246^{\ln\left(\text{n}\right)}\tag{11}$$

Answer 2

1. Rewrite the log in base $\frac{6}{7}$
2. Use $a^{bc} = a^{b}a^{c}$ to be able to simplify the log.