Pls hurry , timed test
1 answer:
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$\implies 2^{2x}-2\cdot5^{2x}-(2^x)(5^x)>0$
let $2^x=a$ and $5^x=b$
So, $a^2-2b^2-ab>0$
divide by $b^2$, ($b^2>0$)
$\implies \left(\frac ab\right)^2-\left(\frac ab\right) -2>0$
this is a quadratic, in $\left(\frac ab\right)$, let it be $x$
So, $x^2-x-2>0$
Can you simplify it now?
Solve equation [1] for the variable u
[1] u = v
// Plug this in for variable u in equation [2]
[2] 8•(v ) - 4v = -32
[2] 4v = -32
// Solve equation [2] for the variable v
[2] 4v = - 32
[2] v = - 8
// By now we know this much :
u = v
v = -8
// Use the v value to solve for u
u = (-8) = -8
Solution : {u , v } = { -8,-8}
[1] u = v
// Plug this in for variable u in equation [2]
[2] 8•(v ) - 4v = -32
[2] 4v = -32
// Solve equation [2] for the variable v
[2] 4v = - 32
[2] v = - 8
// By now we know this much :
u = v
v = -8
// Use the v value to solve for u
u = (-8) = -8
Solution : {u , v } = { -8,-8}