Question

প্রমাণ কর যে: $\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot6^{p}}{3^{p-2}\cdot6^{2p+2}\cdot10^{p}\cdot15^{q}}=\dfrac12$

Answer 1

Right Hand Side,
$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot6^{p}}{3^{p-2}\cdot6^{2p+2}\cdot10^{p}\cdot15^{q}}$

$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot\left(2\times3\right)^{p}}{3^{p-2}\cdot\left(2\times3\right)^{2p+2}\cdot\left(2\times5\right)^{p}\cdot\left(3\times5\right)^{q}}$

$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot2^{p}\cdot3^{p}}{3^{p-2}\cdot2^{2p+2}\cdot3^{2p+2}\cdot2^{p}\cdot5^{p}\cdot3^{q}\cdot5^{q}}$

$=\dfrac{2^{\left(2p+1\right)+p}\cdot3^{\left(2p+q\right)+p}\cdot5^{p+q}}{2^{\left(2p+2\right)+p}\cdot3^{\left(p-2\right)+\left(2p+2\right)+q}\cdot5^{p+q}}$

$=\dfrac{2^{2p+1+p}\cdot3^{2p+q+p}\cdot5^{p+q}}{2^{2p+2+p}\cdot3^{p-2+2p+2+q}\cdot5^{p+q}}$

$=\dfrac{2^{3p+1}\cdot3^{3p+q}\cdot5^{p+q}}{2^{3p+2}\cdot3^{3p+q}\cdot5^{p+q}}$

$=\dfrac{2^{3p+1}}{2^{3p+2}}\cdot\frac{3^{3p+q}}{3^{3p+q}}\cdot\frac{5^{p+q}}{5^{p+q}}$

$=\dfrac{2^{3p+1}}{2^{3p+2}}\cdot1\cdot1$

$=\dfrac{2^{3p+1}}{2^{3p+2}}$

$=2^{\left(3p+1\right)-\left(3p+2\right)}$

$=2^{3p+1-\left.3p-2\right)}$

$=2^{-1}$

$=\frac12$

$=$ Right Hand Side [Proved]

Comments

Rules Applied:

• $(ab)^{m}=a^{m}\cdot b^{n}$
• $a^{m} \cdot a^{n}=a^{m+n}$
• $\frac{a^{m}}{a^{n}}=a^{m-n}$
• $a^{-1}=\frac{1}{a}$

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Why? $a^{-1}=\frac{1}{a}$

$\because a^{-1}=a^{0-1}=\frac{a^0}{a^1}=\frac{1}{a}$

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Why? $a^0=1$

$\because a^0=a^{m-m}=\frac{a^m}{a^m}=1$