How to solve this ( and general )logarithmic Equation $$ 4^{x+1} - 6^x - 2*9^{x+1} = 0$$
I recently stumbled across this logarithmic Equation and really I have no clue how to solve this. Also , please provide a tactical way to approach such questions 
 A: HINT: You can make $a=2^x$ and $b=3^x$, so that your equation becomes
$$4a^2-ab-18b^2=0$$
$$(4a-9b)(a+2b)=0$$
A: Divide by $9^x$ and then you get a quadratic in $(2/3)^x$.
A: Hint: dividing by $2^{2x}$ will result in quadratic equation for $u=\frac{3}{2}$.
A: $$4^{x+1} - 6^x - 2 \cdot 9^{x+1} = 0\\
2^{2x+2} - 2^x \cdot 3^x - 2 \cdot 3^{2x+2} = 0$$
Now, divide everything by $2^{2x}$:
$$4^{x+1} - 6^x - 2 \cdot 9^{x+1} = 0\\
2^{2} - \frac{3^x}{2^x} - \frac{2 \cdot 3^{2x+2}}{2^{2x}} = 0 \\
4 - \left(\frac{3}{2}\right)^x - 2\cdot9 \left(\frac{3}{2}\right)^{2x} = 0$$
Let's pose $y = \left(\frac{3}{2}\right)^x.$ Then, your equation can be rewritten as follows:
$$4 - y - 18y^2 = 0 \Rightarrow y = -\frac{1}{2} ~\text{or}~ y = \frac{4}{9}.$$
Of course, we need to discard the negative solution, since $y = \left(\frac{3}{2}\right)^x$ must be positive by definition.
Finally:
$$\left(\frac{3}{2}\right)^x = \frac{4}{9} \Rightarrow 
x\log\left(\frac{3}{2}\right) = \log\left(\frac{4}{9}\right) \Rightarrow \\
x = \frac{\log\left(\frac{4}{9}\right)}{\log\left(\frac{3}{2}\right)}  \Rightarrow
x = -\frac{\log\left(\frac{9}{4}\right)}{\log\left(\frac{3}{2}\right)} \Rightarrow\\
x = -2\frac{\log\left(\frac{3}{2}\right)}{\log\left(\frac{3}{2}\right)} \Rightarrow
x = -2.$$
A: Hint:
$$2^{x+1}\times 2^{x+1} -2^x\times 3^x-2\cdot3^{x+1}\times 3^{x+1}=0$$
$$2\cdot 2^x\cdot 2^x\cdot 2-2^x\cdot 3^x-2\cdot 3^x\cdot3\cdot3^x\cdot 3=0$$
$$a:=2^x,\qquad b:=3^x$$
$$4\cdot a^2-a\cdot b-18\cdot b^2=0$$
$$(a+2b)(4a-9b)=0$$
