Question

Prove that
2^30+2^29+2^28/2^31+2^30-2^29=7/10

Answer 1

[tex]\text{LHS} \\ \\ =\frac{2^{30}+2^{29}+2^{28}}{2^{31}+2^{30}-2^{29}}\\ \\ =\frac{2^{28+2}+2^{28+1}+2^{28}}{2^{29+2}+2^{29+1}-2^{29}}\\ \\ \text{Applying }a^{m+n} =a^m\times a^n:\\ \\ =\frac{2^{28}\times2^2+2^{28}\times2^1+2^{28}}{2^{29}\times2^2+2^{29}\times2^1-2^{29}}\\ \\ \text{Factor out }x^{28} \text{from numerator and }x^{29} \text{ from denominator:}\\ \\ = \frac{2^{28}(2^2+2^1+1)}{2^{29}(2^2+2^1-1)}\\ \\ \text{Simplify further:}\\ \\ = \frac{4+2+1}{2(4+2-1)} \\ \\ = \frac{7}{2(5)} \\ \\ [/tex]

[tex]= \frac{7}{10}\\ \\ \text{=RHS (Proved)} [/tex]

Answer 2

Answer:

Required to prove,

$\frac{2^{30}+2^{29}+2^{28}}{2^{31}+2^{30}-2^{29}} = \frac{7}{10}$

Recall the concepts:

From the law of exponents

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

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

Solution:

LHS = $\frac{2^{30}+2^{29}+2^{28}}{2^{31}+2^{30}-2^{29}}$

Taking out  $2^{28}$  from the terms of the numerator and $2^{29}$ from the terms of the denominator we get

$\frac{2^{30}+2^{29}+2^{28}}{2^{31}+2^{30}-2^{29}}$ = $\frac{2^{28}(2^2 + 2 +1)}{2^{29}(2^2 + 2 -1)}$ = $\frac{2^{28}X 7}{2^{29}X 5}$

By applying equation (1) we get,

$\frac{2^{28}X 7}{2^{29}X 5}$  = $\frac{2^{-1}X 7}{ 5}$

Applying equation (2) we get

= $\frac{7}{2 X 5}$ = $\frac{7}{10}$ = RHS

∴$\frac{2^{30}+2^{29}+2^{28}}{2^{31}+2^{30}-2^{29}} = \frac{7}{10}$

Hence proved

#SPJ2