Question

If 2^m+n / 2^n-m = 16 & a = 2^1/10 then a^2m+n-p / (am-2n+2p)^-1

Answer 1

Answer:

$\frac{(a^{2m+n-p})^2}{(a^{m -2n + 2p})^{-1}} = 2$

Step-by-step explanation:

If 2^m+n / 2^n-m = 16 & a = 2^1/10 then a^2m+n-p / (am-2n+2p)^-1

$\frac{2^{m+n}}{2^{n-m}} = 16\\ \\\implies 2^{m + n - n + m} = 2^4 \\ \\\implies 2^{2m} = 2^4 \\ \\\implies 2m = 4$

=> m = 2

$\frac{(a^{2m+n-p})^2}{(a^{m -2n + 2p})^{-1}} \\ \\\implies a^{2m+n-p} \times a^{2m+n-p} \times a^{m -2n + 2p} \\ \\\implies a^{2m + n - p + 2m + n - p + m - 2n + 2p} \\ \\\implies a^{5m} \\ \\m =2\\ \\\implies a^{10}$

$a = 2^{\frac{1}{10}}\\ \\\implies a^{10} = (2^{\frac{1}{10}})^{10} = 2$

Answer 2

Step-by-step explanation:

If 2^m+n / 2^n-m = 16 & a = 2^1/10 then a^2m+n-p / (am-2n+2p)^-1

\begin{gathered}\frac{2^{m+n}}{2^{n-m}} = 16\\ \\\implies 2^{m + n - n + m} = 2^4 \\ \\\implies 2^{2m} = 2^4 \\ \\\implies 2m = 4 \\\end{gathered}

2

n−m

2

m+n

=16

⟹2

m+n−n+m

=2

4

⟹2

2m

=2

4

⟹2m=4

=> m = 2

\begin{gathered}\frac{(a^{2m+n-p})^2}{(a^{m -2n + 2p})^{-1}} \\ \\\implies a^{2m+n-p} \times a^{2m+n-p} \times a^{m -2n + 2p} \\ \\\implies a^{2m + n - p + 2m + n - p + m - 2n + 2p} \\ \\\implies a^{5m} \\ \\m =2\\ \\\implies a^{10} \\\end{gathered}

(a

m−2n+2p

)

−1

(a

2m+n−p

)

2

⟹a

2m+n−p

×a

2m+n−p

×a

m−2n+2p

⟹a

2m+n−p+2m+n−p+m−2n+2p

⟹a

5m

m=2

⟹a

10

\begin{gathered}a = 2^{\frac{1}{10}}\\ \\\implies a^{10} = (2^{\frac{1}{10}})^{10} = 2\end{gathered}

a=2

10

1

⟹a

10

=(2

10

1

)

10

=2