Solving Logs different bases?

Question

I do not understand how $\log_2(x) + \log_4(x) = \log_2({x^{3/2}})$

Where does $^{3/2}$ come from? Naming the rules and steps would be helpful.

Answer 1

$$\log_2x+\log_4x=\log_2x+\frac{\log_2x}{\log_24}=\log_2x+\frac12\log_2x=\frac32\log_2x=\log_2x^{3/2}$$

Answer 2

First rearrange your equation: $$\log_2(x) + \log_4(x) = \log_2({x^{3/2}})$$ Now since $\log_2(x^c)=c\log_2(x)$ we have: $$\log_2(x) + \log_4(x) = \frac{3}{2}\log_2(x)$$ Therefore $$\log_4(x)=\frac{3}{2}\log_2(x)-\log_2(x)=\frac{1}{2}\log_2(x)\tag{1}$$

Now let $a=\log_2(x)$. Then $x=2^a$, and so $\log_4(x)=\log_4(2^a)=a\log_4(2)$, hence $$a=\frac{\log_4(x)}{\log_4(2)}=\log_2(x)\tag{2}$$ or $$\log_4(x)=\log_4(2)\log_2(x)\tag{3}$$ But $\log_4(2)=b$ implies $2=4^b$, and so $b=\frac{1}{2}$. Therefore plugging this into (3) gives $\log_4(x)=\frac{1}{2}\log_2(x)$ as required by (1).