Given that $2^{a} + 2^{a+1} + 2^{a+2} + 2^{a+3} = 32$, find the value of a. Answer (a) a = ... Solve the equation $25^{x+2} \times 125 \div 5^{-x} = 1$. Answer (b) x = ...

1. Factor out the common term $2^a$ from the left-hand side.

$2^a+2^{a+1}+2^{a+2}+2^{a+3}=2^a(1+2+4+8)$
So,
$2^a(15)=32$

2. Solve for $a$.

From
$15\cdot 2^a=32,$
$2^a=\frac{32}{15}.$
Hence
$a=\log_2\left(\frac{32}{15}\right).$

1. Rewrite all powers using base 5.

Since $25=5^2$ and $125=5^3$,
$25^{x+2}\times125\div5^{-x}=1$
$\Rightarrow (5^2)^{x+2}\times5^3\div5^{-x}=1.$

2. Simplify the left-hand side using laws of indices.

$(5^2)^{x+2}=5^{2x+4}$
So
$5^{2x+4}\times5^3\div5^{-x}=5^{2x+4+3-(-x)}=5^{3x+7}.$

3. Equate the exponent to zero and solve for $x$.

Since $1=5^0$,
$5^{3x+7}=5^0.$
Hence,
$3x+7=0$
$3x=-7$
$x=-\frac{7}{3}.$