Rules of exponents

Algebra Level 2

Solve for x x .

1 27 ⋅ 3 59 ⋅ 1 243 ⋅ 2 7 x = 1 3 ⋅ 3 x \large{\dfrac{1}{27} \cdot 3^{59} \cdot \dfrac{1}{243} \cdot 27^x=\dfrac{1}{3} \cdot 3^x}

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The answer is -26.

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1 solution

1 27 ⋅ 3 59 ⋅ 1 243 ⋅ 2 7 x = 1 3 ⋅ 3 x \dfrac{1}{27} \cdot 3^{59} \cdot \dfrac{1}{243} \cdot 27^x=\dfrac{1}{3} \cdot 3^x

1 3 3 ⋅ 3 59 ⋅ 1 3 5 ⋅ 3 3 x = 3 x − 1 \dfrac{1}{3^3} \cdot 3^{59} \cdot \dfrac{1}{3^5} \cdot 3^{3x}=3^{x-1}

3 51 ⋅ 3 3 x = 3 x − 1 3^{51} \cdot 3^{3x}=3^{x-1}

3 51 + 3 x = 3 x − 1 3^{51+3x}=3^{x-1}

Since they have the same base, we have

51 + 3 x = x − 1 51+3x=x-1

2 x = − 52 2x=-52

x = − 26 x=-26

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