log equation hard to solve, logarithm with a polynomial

Question

I could not solve this logarithmic equation, it proved difficult to me, just hints or full solution if you can please: $$\log_4(3\cdot4^{x+1}-8)=2x+1$$

Answer 1

We can solve this equation by bringing it to a quadratic form. $$ \log_2(3\cdot 2^{x+1}-4)=2x+1\\ $$ $$ \Leftrightarrow 6\cdot 2^x -4=2^{2x+1} $$ $$ \Leftrightarrow 6\cdot 2^x -4=2\cdot (2^{x})^2 $$

Substitute z=2^x to get a quadratic equation: $$ 6\cdot z -4=2\cdot z^2 $$ $$ \Leftrightarrow z^2 -3z+2=0 $$ with the solutions $z=1$ and $z=2$. Thus, with $z=2^x$ we get the solutions $x=1$ and $x=0$.

As somehow the question got changed, here the solution to the new one: $$ \log_4(3\cdot 4^{x+1}-8)=2x+1\\ $$ $$ \Leftrightarrow 12\cdot 4^x -8=4^{2x+1} $$ $$ \Leftrightarrow 12\cdot 4^x -8=4\cdot (4^{x})^2 $$

Substitute $z=4^x$ to get a quadratic equation: $$ 12\cdot z -8=4\cdot z^2 $$ $$ \Leftrightarrow z^2 -3z+2=0 $$ with again the solutions $z=1$ and $z=2$. Thus, with $z=4^x$ we get the solutions $x=0.5$ and $x=0$.

Answer 2

$\log_2(3\cdot2^{x+1}-4)=2x+1 \iff 6 \cdot2^x-4=2(2^x)^2.$

Let $t=2^x$. This gives $t^2-3t+2=0.$

Can you proceed ?

Answer 3

$$ \log_4(3\cdot 4^{x+1}-8)=2x+1\\ $$ $$ \rightarrow 12\cdot 4^x-8=4^{2x+1} $$ $$ \rightarrow 12\cdot 4^x -8=4\cdot (4^{x})^2 $$

Substitute $z=4^x$ to get a quadratic equation: $$ 12\cdot z -8=4\cdot z^2 $$ $$ \rightarrow z^2-3z+2=0 $$ Thus $z=1$, $z=2$. You can now calculate the value of x.