What are the exact steps to produce $k = \log_{3/2}n $ from $(\frac{2}{3})^kn = 1$?

Question

What are the procedures to derive $k = \log_{3/2}n $ from $(\frac{2}{3})^kn = 1$? Is there a well-known formula?

Answer 1

We can write $$\left(\frac{2}{3}\right)^k=\frac{1}{n}$$ so $$\left(\frac{3}{2}\right)^k=n$$ so $$k\ln \left(\frac{3}{2}\right)=\ln(n)$$

Answer 2

Pass $n$ to the right and take $log_{\frac{3}{2}}$ in both sides. Now using that $log(a^b) = b \log(a)$ you get on the left $k *log_{\frac{3}{2}}(\frac{2}{3})$ which is $-k$. On the right you have $log_{\frac{3}{2}}(\frac{1}{n})$ which is $-log_{\frac{3}{2}}(n)$. So you finally get $k=log_{\frac{3}{2}}(n)$

Answer 3

It's just definition.

$(\frac 23)^k n = 1$ so

$(\frac 23)^k = \frac 1n$ so

$(\frac 32)^k = n$ so

By definition, $\log_{\frac 32} n = k$.

The formulas $\log_b n = - \log_{\frac 1b} n$ (because $b^k = (\frac 1b)^{-k}$) and $\log_b \frac 1n = -\log_b n$ (ditto) might be useful.

But mustly you should know the definition that $\log_b n = k \iff b^k = n$. Technically that isnt a "formula" (because it's a definition) but it may be thought of as one.

Answer 4

$k = \log_{3/2} n$ $\Rightarrow$ $\left( \frac{3}{2} \right)^k = n$ (by definition of logarithm) $\Rightarrow$ $1 = \left( \frac{3}{2} \right)^{-k}n$ $\Rightarrow$ $n\left( \frac{2}{3} \right)^k = 1$

Answer 5

Manipulate the original equation a bit:

$$\left(\frac{2}{3}\right)^k n = 1 \iff \left(\frac{2}{3}\right)^k = \frac{1}{n} \iff \left(\frac{3}{2}\right)^k = n$$

Now, just use the definition of a logarithm: $a^b = c \iff \log_a c = b$. This gets the desired result: $k = \log_{\frac{3}{2}} n$.