I am trying to solve this exponential inequality $$ \frac{4}{7}\sqrt{4 - 4 \cdot 7^{x}+7^{2x}} > 7^{2x-1}-4 \cdot 7^{x-1}+1 $$
My solution is $\mathbb{R}$, since $a=2$ and union of the two possible solutions is $\mathbb{R}$.
Can anyone confirm this or correct me? Thanks!
On
Put $y=7^{x}$ and note that $\sqrt {4-4y+y^{2}}=|2-y|$. Hence the inequality is $\frac 4 7 |2-y| >\frac 1 7 y^{2}-\frac 4 7 y+1$. This is certainly false for large values of $y$ , hence for large values of $x$. So your answer is not correct. Now try solving the inequality $4 |2-y| > y^{2}- 4 y+7$.
Multiply by $7$ and rewrite the inequality $\;\frac{4}{7}\sqrt{4-4\cdot 7^{x}+7^{2x}} > 7^{2x-1}-4\cdot 7^{x-1}+1\;$ as $$4\cdot \sqrt{(7^x-2)^2} > (7^x-2)^2+3.$$ Set $\;t=|7^x-2|$ and solve the quadratic inequality $$4t>t^2+3.$$ We get $\;t\in(1,3)$ or $$1<|7^x-2|<3.$$ This gives $$-1<7^x<1\quad \text{or}\quad 3<7^x<5.$$ I think you can finish from this.