Resolved: "(x4y5)14(x8y5)15=xj5yk4 In the equation above, j and k..."

College
Advanced Math
Upper Level Math

kuCAu

Answered question

2021-03-08

${(x^{4} y^{5})}^{\frac{1}{4}} {(x^{8} y^{5})}^{\frac{1}{5}} = x^{\frac{j}{5}} y^{\frac{k}{4}}$
In the equation above, j and k are constants. If the equation is true for all positive real values of x and y, what is the value of $j - k$ ?
A)3
B)4
C)5
D)6

Answer & Explanation

pattererX

Skilled2021-03-09Added 95 answers

Step 1
The given equation is, ${(x^{4} y^{5})}^{\frac{1}{4}} {(x^{8} y^{5})}^{\frac{1}{5}} = x^{\frac{j}{5}} y^{\frac{k}{4}}$
Step 2
Assume that the abive equation is true for all positive real values of x and y.
Now obtain the values of constants j and k as shown below.
${(x^{4} y^{5})}^{\frac{1}{4}} {(x^{8} y^{5})}^{\frac{1}{5}} = x^{\frac{j}{5}} y^{\frac{k}{4}}$
${(x^{4})}^{\frac{1}{4}} {(y^{5})}^{\frac{1}{4}} {(x^{8})}^{\frac{1}{5}} {(y^{5})}^{\frac{1}{5}} = x^{\frac{j}{5}} y^{\frac{k}{4}}$
$(x y^{\frac{5}{4}}) (x^{\frac{8}{5}}) = x^{\frac{j}{5}} y^{\frac{k}{4}}$
$(x^{1 + \frac{8}{5}}) (y^{1 + \frac{5}{4}}) = x^{\frac{j}{5}} y^{\frac{k}{4}}$
$x^{\frac{13}{5}} y^{\frac{9}{4}} = x^{\frac{j}{5}} y^{\frac{k}{4}}$
Step 3
Equate the powers and obtain the values of j and k as follows.
$\frac{j}{5} = \frac{13}{5}$
$j = 13$
$\frac{k}{4} = \frac{9}{4}$
$k = 9$
Now compute the difference $j - k$ as shown below.
$j - k = 13 - 9$
$= 4$

Therefore, the correct option is B.

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