Question

solve this - find the value of m

Answer 1

Answer :-

m = 1

Given to find the value of m :-

$\dfrac{(16)^{2m+1}(64)^5}{(256)^2\times 4} = (256)^{3m}$

Formulae used :-

Here some laws of exponents can be used ,

$(a^m)^n = a^{mn}$

$a^m \times a^n = a^{m+n}$

If bases are equal then powers also should be equal

Solution :-

As we know that ,

$\begin{gathered}\begin{array}{c | c} \underline2 &\underline{256} \\ \underline2&\underline{128} \\ \underline2 &\underline{64}\\ \underline2&\underline{32} \\ \underline2&\underline{16} \\ \underline2&\underline8 \\ \underline2&\underline4\\ \underline2&\underline2 \\ &1\end{array}\end{gathered}$

256 = 2⁸

$\begin{gathered}\begin{array}{c | c} \underline2 &\underline{64} \\ \underline2&\underline{32} \\ \underline2 &\underline{16}\\ \underline2&\underline{8} \\ \underline2&\underline{4} \\ \underline2&\underline2 \\ &\underline1 \\ &\end{array}\end{gathered}$

64 =  2⁶

$\begin{gathered}\begin{array}{c | c} \underline2 &\underline{16} \\ \underline2&\underline{8} \\ \underline2 &\underline{4}\\ \underline2&\underline{2} \\ &\underline{1} \\ &\end{array}\end{gathered}$

16 = 2⁴

So, Lets simplify the given question by substituting these values

$\dfrac{(16)^{2m+1}(64)^5}{(256)^2\times 4} = (256)^{3m}$

$\dfrac{(2^4)^{2m+1}(2^6)^5}{(2^8)^2\times 2^2 } = (2^8)^{3m}$

$\dfrac{2^{4(2m+1)}2^{6\times5}}{2^{8\times 2}\times 2^2 } = 2^{8\times 3m}[(a^m)^n = a^{mn}]$

$\dfrac{2^{8m+4}\times 2^{30}}{2^{16}\times 2^2}= 2^{24m}$

$\dfrac{2^{8m+4+30}}{2^{16+2}}= 2^{24m} [ a^m \times a^n = a^{m+n}]$

$\dfrac{2^{8m+34}}{2^{18}} = 2^{24m}$

Do cross multiplication

$2^{8m+34} = 2^{18} \times 2^{24m}$

$2^{8m+34} = 2^{18+24m}$

As also we know that ,

If bases are equal powers also should be equal

So,

$8m +34 = 18+24m$

$8m -24m = 18-34$

$-16m= -16$

$m = 1$

So, the  value of m is 1