Question

81/16^-5/4*25/9^-5/2

Answer 1

Answer: - We will get the value of the equation after solving it = (2/5)^5 OR 32/3125.

Detailed solution: -

Given: -

The given equation needs to be solved or evaluate it.

The given equation is: -

$(\frac{81}{16})^\frac{-5}{4} * (\frac{25}{9})^\frac{-5}{2}$

We need to solve it, we will solve it step wise.

First, we will take out the LCM of the numbers 81, 16, 25 and 9.

We will get,

The LCM of 81 = 3×3×3×3

The LCM of 16 = 2×2×2×2

The LCM of 25 = 5×5

The LCM of 9 = 3×3

Therefore,

Now we will write the LCM of every number and write then in the power and base form.

Then the denominator of the power will get cancel with the power from LCM.

So,

$(\frac{81}{16})^\frac{-5}{4} * (\frac{25}{9})^\frac{-5}{2} \\\\ =(\frac{3*3*3*3}{2*2*2*2})^\frac{-5}{4}*(\frac{5*5}{3*3})^\frac{-5}{2} \\\\ =( \frac{3}{2})^4^*^\frac{-5}{4} *(\frac{5}{3})^2^*^\frac{-5}{2}$

4 will be cancelled with the denominator 4 and 2 will be cancelled with the denominator 2.

Then we will get,

$=(\frac{3}{2})^-^5*(\frac{5}{3})^-^5$

After the reciprocal of the fractions given, the power will be in its positive form.

Therefore,

$=(\frac{2}{3})^5*(\frac{3}{5})^5\\ \\=\frac{2*2*2*2*2}{3*3*3*3*3}* \frac{3*3*3*3*3}{5*5*5*5*5}$

The line of 3 in the numerator will be cancelled by the line of 3 from the denominator, i.e., we will get,

$=\frac{2*2*2*2*2}{5*5*5*5*5}\\ \\=\frac{32}{3125}$

OR

$=(\frac{2}{5}) ^5$

Therefore,

We will get the value of the equation after solving it = (2/5)^5 OR 32/3125

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