Algebraically, how do you determine the intersection point of the functions $y = {(\frac{1}{2})}^{- 14 x + 1}$ $y=(1/2)^(-14x+1)$ and $y = 8^{2 x + 1}$ $y=8^(2x+1)$ ?

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Answer 1 · 2017-01-04T03:17:02

$(\frac{1}{2}, 64)$ $(1/2, 64)$ is the point of intersection.

Solve through substitution.

$y = {(\frac{1}{2})}^{- 14 x + 1} \to {(\frac{1}{2})}^{- 14 x + 1} = 8^{2 x + 1}$ $y= (1/2)^(-14x+ 1) -> (1/2)^(-14x + 1) = 8^(2x + 1)$

Write in equivalent bases.

${(2^{- 1})}^{- 14 x + 1} = {(2^{3})}^{2 x + 1}$ $(2^-1)^(-14x + 1) = (2^3)^(2x + 1)$

$2^{14 x - 1} = 2^{6 x + 3}$ $2^(14x - 1) = 2^(6x + 3)$

Eliminate the bases since they're now equivalent.

$14 x - 1 = 6 x + 3$ $14x - 1 = 6x + 3$

$14 x - 6 x = 3 + 1$ $14x - 6x = 3 + 1$

$8 x = 4$ $8x= 4$

$x = \frac{1}{2}$ $x= 1/2$

Now input this value of $x$ $x$ into one of the equations:

$y = 8^{2 (\frac{1}{2}) + 1}$ $y= 8^(2(1/2) + 1)$

$y = 8^{2}$ $y = 8^2$

$y = 64$ $y = 64$

Hopefully this helps!

Algebraically, how do you determine the intersection point of the functions y=(12)−14x+1y=(1/2)^(-14x+1) and y=82x+1y=8^(2x+1)?