If \( a^{n+n} \times b^{m-n}=6^{4} \) and \( a^{n} \times b^{n}=16 \) find The value of \( a \) and
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To solve the equations \( a^{n+n} \times b^{m-n} = 6^{4} \) and \( a^{n} \times b^{n} = 16 \), we can rewrite them in a more manageable form. Starting with the second equation, \( a^{n} \times b^{n} = 16 \), we can express this as: \[ ab = 16^{1/n} \] For the first equation, we can replace \( a^{n+n} \) with \( (a^n)^2 \) and write it as: \[ (a^n)^2 \times b^{m-n} = 6^{4} \] Now, if we let \( a^n = x \) and \( b^n = y \), substituting gives: \[ x^2 \times b^{m-n} = 6^4 \] With both equations transformed, we can solve for \( a \) and \( b \) in terms of \( x \) and \( y \). We find the relationships between \( x \) and \( y \) using the equations: 1. \( xy = 16 \) 2. \( x^2 \times (y^{m/n}) = 6^4 \) Substituting \( y = \frac{16}{x} \) into the first transformed equation and solving gives the values for \( a \) and \( b \) based on the determined values of \( x \) and \( y \).