Given that $\log_6 2 = .387$ ,find $\log_6 9$

Question

Given that $\log_62 = .387$. Find $\log_6 9$.

I don't know proper method to solve this problem . Please help me.

Answer 1

Let $\log_62=a$.

Hence,

$$\log_62=\frac{1}{1+\log_23}$$ and $$\log_69=\frac{2\log_23}{1+\log_23}.$$

Thus, $$\log_23=\frac{1-a}{a}$$ and $$\log_69=\frac{\frac{2(1-a)}{a}}{1+\frac{1-a}{a}}=2(1-a)=1.226...$$

Now we see a best way: $$\log_69=\log_6\frac{36}{4}=\log_636-\log_64=2-2\log_62=...$$

Answer 2

This is probably the most brute-force approach, and it is reasoning like this (but with general variables rather than concrete numbers) that lies behind the formulas in the other answer.

We are after $x$ such that $$ 6^x = 9\\ 2^x3^x = 3^2\\ 2^x = 3^{2-x} $$ We are given that $$ 6^{0.387} = 2\\ 2^{0.387}3^{0.387} = 2^1\\ 2^{0.387-1} = 3^{-0.387}\\ 2^{0.613} = 3^{0.387} $$ Raising the last equation to the power of $x/0.613$ yields $$ 2^{x}= 3^{0.387x/0.613} $$but above we see that $2^x$ at the same time is equal to $3^{x-2}$. This means that we have $3^{x-2} = 3^{0.387x/0.613}$, which can be solved relatively easily.

Answer 3

$\log_6(2) = .387 =:a$

$ \rightarrow$:

$6^a = (2×3)^a =2^a3^a = 2;$

$3^a = 2^{1-a}, $

now square both sides:

$9^a = 2^{2(1-a)}.$

Taking $\log_6$ of both sides:

$a\log_6(9) = 2(1-a)\log_6(2).$

Recall: $a = \log_6(2)$:

$\log_6(9) =2 (1-a).$