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jmc

algebra intermediate

Problem

The expression

lo g_{y^{6}} x \cdot lo g_{x^{5}} y^{2} \cdot lo g_{y^{4}} x^{3} \cdot lo g_{x^{3}} y^{4} \cdot lo g_{y^{2}} x^{5}

can be written as

a lo g_{y} x

for what constant

a

?

Solution

First we use the formula

lo g_{a} b = \frac{l o g _{c} b}{l o g _{c} a}

. The given expression becomes

lo g_{y^{6}} x \cdot lo g_{x^{5}} y^{2} \cdot lo g_{y^{4}} x^{3} \cdot lo g_{x^{3}} y^{4} \cdot lo g_{y^{2}} x^{5} = \frac{lo g x}{lo g y ^{6}} \cdot \frac{lo g y ^{2}}{lo g x ^{5}} \cdot \frac{lo g x ^{3}}{lo g y ^{4}} \cdot \frac{lo g y ^{4}}{lo g x ^{3}} \cdot \frac{lo g x ^{5}}{lo g y ^{2}}

Next we use the formula

a lo g_{b} x = lo g_{b} x^{a}

. We get

\frac{lo g x}{lo g y ^{6}} \cdot \frac{lo g y ^{2}}{lo g x ^{5}} \cdot \frac{lo g x ^{3}}{lo g y ^{4}} \cdot \frac{lo g y ^{4}}{lo g x ^{3}} \cdot \frac{lo g x ^{5}}{lo g y ^{2}} = \frac{lo g x}{6 lo g y} \cdot \frac{2 lo g y}{5 lo g x} \cdot \frac{3 lo g x}{4 lo g y} \cdot \frac{4 lo g y}{3 lo g x} \cdot \frac{5 lo g x}{2 lo g y} = \frac{120 lo g x}{720 lo g y} = \frac{lo g x}{6 lo g y} = \frac{1}{6} lo g_{y} x .

Therefore,

a = \frac{1}{6}

.

Final answer

\frac16