Answer is 3/2 to this question:
But how? And why?
Asked
Active
Viewed 42 times
0
-
Do you know how to use log properties? – Sean Roberson Nov 28 '21 at 18:06
-
In particular $\log_{a}\left(a^b\right)=b$ – Henry Nov 28 '21 at 18:07
-
I know all the log properties @SeanRoberson – Yağız Alp Ersoy Nov 28 '21 at 18:08
-
1It holds $2=4^{1/2}$. Does it help you? – Jochen Nov 28 '21 at 18:17
-
Suppose that $(12)^r = s$. Then, since $(12) = (2^2 \times 3)$ you have that $2^{(2r)} \times 3^{(r)} = s.$ Notice that the exponents that are assigned to the bases of $(2)$ and $(3)$ in the expression $2^{(2r)} \times 3^{(r)} = s$ are in the proportion of $[2r ~: ~r].$ – user2661923 Nov 28 '21 at 18:27
-
That's it! Thank you! – Yağız Alp Ersoy Nov 28 '21 at 18:42
1 Answers
0
That's a pretty easy question, you just need to know how to use some log properties.
Let's get started.
$$\log_{12} ({2^3 \cdot 3^\frac{3}{2}})$$
$$\log_a (p\cdot q) = \log_a p + \log_a q$$
$$\log_{12} ({2^3 \cdot 3^\frac{3}{2}}) = \log_{12} ({2^3}) + \log_{12} {(3^\frac{3}{2})}$$
$$\log_a (p^q) = q\log_a p $$
$$\log_{12} ({2^3 \cdot 3^\frac{3}{2}}) = 3\log_{12} ({2}) + \frac{3}{2}\cdot \log_{12} {(3)}$$
Now we just need to handle the $\log_{12}{(2)}$ and $\log_{12}{(3)}$.
$$\log_{a}{(x)} = \frac{\log_b {(a)}}{\log_{b} {(x)} } $$
Where, $b$ is any base that you want.
Let's handle $\log_{12}{(2)}$ first:
$$\log_{12}{(2)} = \frac{\log 2}{\log 12}$$ Important: $\log_{10} x = \log x$.
Solving $\log 12$:
$$\log 12 = \log (2\cdot 6)$$ Using the log properties that I showed above for multiplication:
$$\log (2\cdot 6) = \log 2 + \log 6$$ $$\log (2\cdot 6) = \log 2 + (\log (2\cdot 3))$$ $$\log (2\cdot 6) = \log 2 + (\log 2 + \log 3)$$ $$\log (12) = 2\log 2 + \log 3$$
So, $\log_{12}{(2)} = \frac{\log 2}{\log 12}$ which is equal to:
$$\log_{12}{(2)} = \frac{\log 2}{2\log 2 + \log 3}$$
Now you are going to do the same to:
$$\log_{12} (3)$$
I will leave the rest as an exercise.
Will
- 19