3 years ago

Solve the exponential equation. 3^4x-1=3^31

Mathematics

1 answer:

Vikentia [17]3 years ago

7 0

Is the equation like this

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Sedbober [7]

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torisob [31]

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$x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$ i.e answer A.

Step-by-step explanation:

This question involves knowing the following power/exponent rule:

$\sqrt[n]{x^m} = x^\frac{m}{n} \\\\so \sqrt[7]{x^2} = x^\frac{2}{7} \\\\and \\\\ \sqrt[5]{y^3} = y^\frac{3}{5} \\$

Next, when a power is on the bottom of a fraction, if we want to move it to the top, this makes the power become negative.

so the y-term, when moved to the top of the fraction, becomes:

$y^{-\frac{3}{5} } \\$

So the answer is: $x^{\frac{2}{7} } y^{-\frac{3}{5} } \\$

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