Solving Logs different bases? 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.
 A: $$\log_2x+\log_4x=\log_2x+\frac{\log_2x}{\log_24}=\log_2x+\frac12\log_2x=\frac32\log_2x=\log_2x^{3/2}$$
A: A good property of logarithmic functions is:
$log_{a^n}b = \frac{1}{n}log_ab$
Proof:
$log_{a^n}b = \frac{logb}{loga^n} = \frac{logb}{n\cdot loga} = \frac{1}{n} \cdot log_ab$
In your example, $log_4(x) = log_{2^2}(x) = \frac{1}{2}log_2(x)$ and now, continue with the common properties of logarithms to solve your problem.
A: 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).
