Question

Find the value of x for 2^(2x+1) +4^(x+1) -384=0

Answer 1

Here,

$\longrightarrow 2^{2x+1}+4^{x+1}=384$

Since $2^2=4,$

$\longrightarrow 2^{2x+1}+(2^2)^{x+1}=384$

Since $(a^m)^n=a^{mn},$

$\longrightarrow 2^{2x+1}+2^{2(x+1)}=384$

$\longrightarrow 2^{2x+1}+2^{2x+2}=384$

$\longrightarrow 2^{2x+1}+2^{2x+1+1}=384$

Since $a^{m+n}=a^m\times a^n,$

$\longrightarrow 2^{2x+1}\times1+2^{2x+1}\times2=384$

Taking $2^{2x+1}$ common in LHS,

$\longrightarrow 2^{2x+1}\left(1+2)=384$

$\longrightarrow 2^{2x+1}\times3=384$

$\longrightarrow 2^{2x+1}=\dfrac{384}{3}$

$\longrightarrow 2^{2x+1}=128$

$\longrightarrow 2^{2x+1}=2^7$

Taking log to the base 2,

$\longrightarrow \log\left(2^{2x+1}\right)=\log\left(2^7\right)$

$\longrightarrow (2x+1)\log2=7\log2$

$\longrightarrow 2x+1=7$

$\longrightarrow 2x=6$

$\longrightarrow\underline{\underline{x=3}}$

Hence 3 is the answer.

Answer 2

$\sf We \ have \implies 2^{2x+1} + 4^{x+1} - 384 = 0$

We have to find x here.

Solution:-

$\sf \implies 2^{2x+1} + 4^{x+1} = 384$

$\sf \implies 2^{2x} \times 2+ 4^{x} \times 4 = 384 \qquad\qquad .....[since, \ a^{1+1} = a \times a]$

$\sf \implies (2^{2})^x \times 2+ 4^{x} \times 4 = 384 \qquad\qquad .....[since, \ a^{bc}=(a^b)^c]$

$\sf \implies 4^x \times 2+ 4^{x} \times 4 = 384$

$\bullet\qquad\bf Let \ 4^x = y$

$\sf \implies y \times 2+ y \times 4 = 384$

$\sf \implies 2y + 4y = 384$

$\sf \implies 6y = 384$

$\sf \implies y = \dfrac{384}{6}=64$

$\bullet \qquad \boxed{ \bf y = 64}$

$\sf\implies 4^x = 64 \qquad\qquad .....[ y = 4^x]$

$\sf \implies 4^x = 4^3$

$\bullet\qquad\bigstar\boxed{\bf x = 3}\bigstar$

Therefore, the value of x is 3.