Question

If a = 2^m and b=2^m+1 show that 8a^3/b^2=2^m+1show that​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• $a=2^{m}$

• $b=2^{m+1}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

$\dfrac{8a^{3} }{b^{2} } =2^{m+1}$

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Given that $a=2^{m}$

→ Hence,

  $a^{3} =({2^{m}}) ^{3}$

 $a^{3} =2^{3m}---(1)$

→ Also given,

  $b=2^{m+1}$

→ Hence,

  $b^{2} =({2^{m+1} })^{2}$

 $b^{2}=2^{2m+2}----(2)$

→ Given LHS is,

   $\dfrac{8a^{3} }{b^{2} }$

→ Substitute equation 1 and 2 in the LHS

  $\dfrac{8a^{3} }{b^{2} } =\dfrac{8\times (2)^{3m} }{(2)^{2m+2} }$

→ We know that 8 = 2³. Substituting in the equation,

 $\dfrac{8a^{3} }{b^{2} } =\dfrac{2^{3} \times (2)^{3m} }{(2)^{2m+2} }$

 $\dfrac{8a^{3} }{b^{2} } =\dfrac{(2)^{3+3m} }{(2)^{2m+2} }$

$\dfrac{8a^{3} }{b^{2} } =(2)^{3+3m-(2m+2)}$

 $\dfrac{8a^{3} }{b^{2} } =(2)^{3+3m-2m-2}$

 $\dfrac{8a^{3} }{b^{2} } =(2)^{m+1}$

= RHS

→ Hence proved.

$\Large{\underline{\underline{\bf{Exponetial\:laws\:used:}}}}$

→ $a^{m}\times a^{n}= a^{m+n}$

→ $\dfrac{a^{m} }{a^{n} } =(a)^{m-n}$

→ $(a^{m})^{n} = a^{mn}$