Question

Difficulty: HardExponents, Radicals, and Algebraic Expressions

If 5x=100y=20z5^x = 100^y = 20^z for non-zero real numbers xx, yy, and zz, which of the following expressions is equal to zz?

  1. xyxy\frac{xy}{x - y}Answer
  2. B
    xyx+y\frac{xy}{x + y}
  3. C
    xyyx\frac{xy}{y - x}
  4. D
    xyx - y
  5. E
    yxy - x

Answer

xyxy\frac{xy}{x - y}
By setting 5x=100y=20z=K5^x = 100^y = 20^z = K, we rewrite each base as a power of KK: 5=K1/x5 = K^{1/x}, 100=K1/y100 = K^{1/y}, and 20=K1/z20 = K^{1/z}. Using the numerical relationship 20=100520 = \frac{100}{5}, we substitute to obtain K1/z=K1/yK1/x=K1y1xK^{1/z} = \frac{K^{1/y}}{K^{1/x}} = K^{\frac{1}{y} - \frac{1}{x}}. Equating exponents gives 1z=1y1x=xyxy\frac{1}{z} = \frac{1}{y} - \frac{1}{x} = \frac{x - y}{xy}. Taking the reciprocal yields the expression stating that zz equals xyxy\frac{xy}{x - y}.

Step-by-Step Solution

1
Set the common exponential expressions equal to a constant KK.
Let 5x=100y=20z=K5^x = 100^y = 20^z = K, where K>0K > 0 and K1K \neq 1.
Equating exponential terms to a shared parameter allows conversion from variable powers to fractional exponents.
2
Express the base numbers 55, 100100, and 2020 in terms of KK.
5=K1/x5 = K^{1/x}, 100=K1/y100 = K^{1/y}, and 20=K1/z20 = K^{1/z}.
Taking the xx-th, yy-th, and zz-th roots of both sides isolates each base number.
3
Relate the base numbers using arithmetic division.
20=1005    K1/z=K1/yK1/x=K1y1x20 = \frac{100}{5} \implies K^{1/z} = \frac{K^{1/y}}{K^{1/x}} = K^{\frac{1}{y} - \frac{1}{x}}.
Dividing powers with the same base subtracts their exponents.
4
Equate the exponents and solve for zz.
1z=1y1x=xyxy    z=xyxy\frac{1}{z} = \frac{1}{y} - \frac{1}{x} = \frac{x - y}{xy} \implies z = \frac{xy}{x - y}.
Taking common denominators and inverting the fraction yields zz in terms of xx and yy.

Key Concept

Relating different bases with variable exponents by setting them equal to a common constant and converting to reciprocal exponents.
Estimated Time:2m 0s
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