Question

4 the power 2 X + 1 + 4 to the power x + 1 equal to 80 then find the value of x​

Answer 1

$\huge\mathfrak\red{Answer}$

${4}^{2x + 1} + {4}^{x + 1} = 80 \\ {4}^{x + x + 1} + {4}^{x + 1} = 80 \\ { 4}^{x} ( {4}^{x + 1} ) + {4}^{x + 1} = 80 \\ {4}^{x + 1} ( {4}^{x} + 1) = 80 \\ {4}^{x} (4)( {4}^{x} + 1) = 80 \\ {4}^{x} ( {4}^{x} + 1) = \frac{80}{4} \\ {4}^{x} ( {4}^{x} + 1) = 20 \\ {4}^{2x} + {4}^{x} = 20 \\ {( {4}^{x} )}^{2} + {4}^{x} = 20 \\ \: \\ let \: {4}^{x} = y \\ {y}^{2} + y = 20 \\ {y}^{2} + y - 20 = 0 \\ {y}^{2} + 5y - 4y - 20 = 0 \\ y(y + 5) - 4(y + 5) = 0 \\ (y - 4)(y + 5) = 0 \\ \\ so \: we \: get \\ {4}^{x} = 4 \\ {4}^{x} = {4}^{1} \\ so \: x = 1 \\ \\ another \: solution \\ {4}^{x} = - 5 \\ take \: log \: on \: both \: side \\ x( log4) = log (- 5) \\ x = \frac{ log( - 5)}{ log4 }$

FORMULA USED.....

$let \: any \: number\\ \\ {a}^{b} = {c}^{d} \\ than \: after \: taking \: log \: on \: both \: side \\ we \: get \\ b \times log(a) = d \times log(c) \\ \\ {4}^{a + b} = {4}^{a} \times {4}^{b}$