Question

$ {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left( {{3^2} \times {5^2}} \right)^2} $

Answer 1

Hint: To solve the expression given in the question we will use exponential laws and BODMAS. If $ x $ and $ y $ are two non-zero integers, then we will use the following laws to resolve the expression.
 $ {x^m} \times {y^m} = {\left( {xy} \right)^m} $
 $ {\left( {{x^m}} \right)^n} = {x^{mn}} $
 $ {x^m} \div {x^n} = {x^{m - n}} $
 $ {x^m} \times {x^n} = {x^{mn}} $

Complete step-by-step answer:
We will apply different laws for resolution of the expression and then we will use BODMAS for the simplification of the problem. BODMAS will define the order of operation that we will perform first. In BODMAS B stands for brackets, O stands for order, D stands for Division, M stands for multiplication, A stands for addition and S stands for Subtraction. We will follow the same order while resolving an expression.

Given:
The given expression is $ {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left( {{3^2} \times {5^2}} \right)^2} $ .
We will have to convert the above equation in the form in which all the exponents are the same.
For this we will use the relation $ {x^m} \times {y^m} = {\left( {xy} \right)^m} $ . This can be expressed as,
 $ \begin{array}{l}
 = {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left( {{3^2} \times {5^2}} \right)^2}\\
 = {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left\{ {{{\left( {3 \times 5} \right)}^2}} \right\}^2}\\
 = {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left\{ {{{\left( {15} \right)}^2}} \right\}^2}
\end{array} $
Now we will use the relation $ {\left( {{x^m}} \right)^n} = {x^{mn}} $ in the above expression to resolve it further.
 $ = {\left( {15} \right)^{50}} \div {\left( {15} \right)^{49}} \times {\left( {15} \right)^4} $
We also know that $ {x^m} \div {x^n} = {x^{m - n}} $ . Hence, we will apply the same relation in the above expression.
 $ \begin{array}{l}
 = {\left( {15} \right)^{50 - 49}} \times {\left( {15} \right)^4}\\
 = {\left( {15} \right)^1} \times {\left( {15} \right)^4}
\end{array} $
Now we will use $ {x^m} \times {x^n} = {x^{mn}} $ in the above expression. So, we get,
 $ \begin{array}{l}
 = {\left( {15} \right)^5}\\
 = 15 \times 15 \times 15 \times 15 \times 15\\
 = 759375
\end{array} $
Therefore, the answer is 759375.

Note: The common mistake that can happen in the above problem is in the resolution of the expression. Exponential laws for division and multiplication are only applicable for the same exponents. Make sure to apply the correct exponential laws in the different conditions