Holiday Problem #5

Algebra Level 3

$\LARGE \sqrt[{x-4}]{ \color{#20A900}5^{ \frac {x}{\sqrt x + 2} } \times (\color{#20A900} {0.2})^{ \frac {4}{\sqrt x + 2} } } = \color{#20A900} {125} \times (\color{#20A900} {0.04})^{ \frac {x-2}{x-4} }$

Solve for $x$

The answer is 9.

1 solution

Curtis Clement
Mar 9, 2015

$(5^{\frac{x}{\sqrt{x} +2}} \times\ 5^{-\frac{4}{\sqrt{x} +2}})^{\frac{1}{x-4}} = 5^{\frac{1}{\sqrt{x} +2}} = 5 ^{3 + \frac{2(2-x)}{x-4}}$ $\Rightarrow\frac{1}{\sqrt{x} +2} = \frac{(3x-12) +4-2x}{x-4} = \frac{x-8}{x-4}$ $\Rightarrow\ x-4 = (x-8)(\sqrt{x} +2)$ Now let $y = \sqrt{x}$ , $y^{2} = x$ : $(y^2 - 8)(y+2) - y^2 - 4 = (y^2 -8)(y+2) - (y+2)(y-2)$ $= (y+2)(y^2 - y -6) = (y+2)^2 (y-3) = 0$ Now $\frac{1}{0}$ is undefined so $\sqrt{x} = y\ne -2$ $\Rightarrow\sqrt{x} = 3 \therefore\ \boxed{x = 9}$ .

Alternatively, you can also proceed like this:

$\frac{1}{\sqrt{x}+2}=\frac{x-8}{x-4}=\frac{x-8}{(\sqrt{x}+2)(\sqrt{x}-2)}\\ \implies \sqrt{x}-2=x-8\\ \implies x-\sqrt{x}-6=0\\ \implies (\sqrt{x}-3)(\sqrt{x}+2)=0\implies \sqrt{x}=3\implies x=9$

The negative solution $\sqrt{x}=(-2)$ is rejected because it makes the original equation undefined.

Prasun Biswas - 6 years, 3 months ago

Holiday Problem #5

The answer is 9.

1 solution

0 pending reports