Logarithm / exponential equation, not sure what to make of this, (simple) Solve for $a:(2 \log_a x)(3 \log_{x^2} 4) = 3$
No idea how to approach this problem other than moving the 2 and the 3 into an exponent..
 A: Use change of base:$\log_xy=\dfrac{\log_ny}{\log_nx}$:
$$(2\log_ax)(3\log_{x^2}4)=(\log_ax^2)(\log_{x^2}4^3)=\frac{\color{red}{\ln x^2}}{\ln a}\cdot\frac{\ln 4^3}{\color{red}{\ln x^2}}=\frac{\ln 4^3}{\ln a}=\log_a4^3=3 \implies a^3=4^3 \implies a=4$$
A: Hints: using the change of base property of logarithms
$$3=\left(2\log_ax\right)\left(3\log_{x^2}4\right)=\color{red}{2}\color{green}{\log_ax}\frac{\log_a64}{\color{red}{2}\color{green}{\log_a x}}=\log_a64\;\ldots$$
A: $(2 \log_a x)(3 \log_{x^2} 4)=\log_a((x^2)^{(\log_{x^2}64)}) = \log_a(64) = 3$ 
A: This answer is full of seemingly useless derivations, but would help you in the future problems. 
There are three interesting results of the base-change property:
First property: $\log _a b = \dfrac{1}{\log _b a }$
Proof. $\log_a b = \dfrac{\log b}{\log a}$ and $\dfrac{1}{\log_b a} = (\log _b a)^{-1} = \left(\dfrac{\log a}{\log b}\right)^{-1} = \dfrac{\log b }{\log a} $
Second property: $\log _{a^m}{b^n} = \dfrac{n}{m}\log _a b$
Proof. As you may be aware of, $\log x^n = n \log x$. 
Hence,$$\log _{a^m}b^n = n \log_{a^ m}b = n \cdot\dfrac{1}{\log _b a^m}= n \cdot \dfrac{1}{m \log_b a} = \dfrac{n}{m} \cdot \dfrac{1}{\log _b a} = \dfrac{n}{m}\log_a b$$
Third property: $a^{\log _x b} = b^{\log _x a}$
Proof. $a^{\frac{\log_a b}{\log_ a x}} = \left(a^{\log _a b}\right)^{\frac{1}{\log _a x}} = b^{\frac{1}{\log_a x}}= b^{\log_x a}$

As for this problem, 
$(2 \log_a x)(3 \log_{x^2} 4) = 3$
$\Rightarrow \log _a x^2 \cdot 3\log_{x^2} 4 = 3$
$\Rightarrow 3\dfrac{\log_{x^2} 4}{\log_{x^2 }a} = 3 $
$\Rightarrow \dfrac{\log_{x^2} 4}{\log_{x^2} a} = 1$
$\Rightarrow \log_a 4 = 1$
$\Rightarrow a^1 = 4$
$\Rightarrow a = 4$
A: $(2 \log_a x)(3 \log_{x^2} 4) = 3\iff( \log_a x^2)( \log_{x^2} 4) = 1$. Then, because of the well known formula $\log_xy\log_y x=1$ one has $$\log_4 x^2\log_{x^2}4=1\Rightarrow\log_4 x^2=\frac{1}{\log_{x^2}4}=\log_a x^2\Rightarrow4^{x^2}=a^{x^2}\iff\left(\frac4a\right)^{x^2}=1$$
Thus (since $g(x)=b^x$ is injective) we have $a=4$.
