Question

How do you solve $ {{\left( \dfrac{1}{4} \right)}^{x}}=64 $ ?

Answer 1

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities We find the relation between the negative exponent and the inverse of the number to find the solution.

Complete step by step answer:
We know the exponent form of the number $ a $ with the exponent being $ n $ can be expressed as $ {{a}^{n}} $ .
In case the value of $ n $ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is $ {{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}} $ .
For our given equation $ {{\left( \dfrac{1}{4} \right)}^{x}}=64 $ , we convert all the given numbers as the power of value 2.
We know that $ 4={{2}^{2}} $ and $ 64={{2}^{6}} $ .
If we take two exponential expressions where the exponents are $ m $ and $ n $ .
Let the numbers be $ {{a}^{m}} $ and $ {{a}^{n}} $ . We take multiplication of these numbers.
The indices get added. So, $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ .
The division works in an almost similar way. The indices get subtracted. So, $ \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} $ .
The left-hand side of the equation $ {{\left( \dfrac{1}{4} \right)}^{x}}=64 $ becomes $ {{\left( \dfrac{1}{4} \right)}^{x}}={{\left( \dfrac{1}{{{2}^{2}}} \right)}^{x}} $ .
Now we use the inverse for negative exponents and get $ {{\left( \dfrac{1}{{{2}^{2}}} \right)}^{x}}={{\left( {{2}^{-2}} \right)}^{x}} $ .
We also have the identity of $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} $ .
Therefore, \[{{\left( {{2}^{-2}} \right)}^{x}}={{2}^{-2x}}\].
We have the final equation where $ {{2}^{-2x}}=64={{2}^{6}} $ .
Now we know that if the bases are equal and power are different as $ {{a}^{m}}={{a}^{n}} $ then $ m=n $ .
For the equation $ {{2}^{-2x}}={{2}^{6}} $ , we get $ -2x=6 $ which gives $ x=\dfrac{6}{-2}=-3 $ .
Therefore, solving the equation $ {{\left( \dfrac{1}{4} \right)}^{x}}=64 $ we get $ x=-3 $ .

Note:
The addition and subtraction for exponents work for taking common terms out depending on the values of the indices.
For numbers $ {{a}^{m}} $ and $ {{a}^{n}} $ , we have $ {{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right) $ .the relation is independent of the values of $ m $ and $ n $ . We need to remember that the condition for $ {{a}^{m}}={{a}^{n}}\Rightarrow m=n $ is that the value of $ a\ne 0,\pm 1 $ .