Solve the following equations for x:
Question:

Solve the following equations for x:

(i) $7^{2 x+3}=1$

(ii) $2^{x+1}=4^{x-3}$

(iii) $2^{5 x+3}=8^{x+3}$

(iv) $4^{2 x}=\frac{1}{32}$

(v) $4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x}$

(vi) $2^{3 x-7}=256$

Solution:

(i)

$7^{2 x+3}=1$

$\Rightarrow 7^{2 x+3}=7^{0}$

$\Rightarrow 2 x+3=0$

$\Rightarrow 2 x=-3$

$\Rightarrow x=-\frac{3}{2}$

(ii)

$2^{x+1}=4^{x-3}$

$\Rightarrow 2^{x+1}=\left(2^{2}\right)^{x-3}$

$\Rightarrow 2^{x+1}=\left(2^{2 x-6}\right)$

$\Rightarrow x+1=2 x-6$

$\Rightarrow x=7$

(iii)

$2^{5 x+3}=8^{x+3}$

$\Rightarrow 2^{5 x+3}=\left(2^{3}\right)^{x+3}$

$\Rightarrow 2^{5 x+3}=2^{3 x+9}$

$\Rightarrow 5 x+3=3 x+9$

$\Rightarrow 2 x=6$

$\Rightarrow x=3$

(iv)

$4^{2 x}=\frac{1}{32}$

$\Rightarrow\left(2^{2}\right)^{2 x}=\frac{1}{2^{5}}$

$\Rightarrow 2^{4 x} \times 2^{5}=1$

$\Rightarrow 2^{4 x+5}=2^{0}$

$\Rightarrow 4 x+5=0$

$\Rightarrow x=-\frac{5}{4}$

(v)

$4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x}$

$\Rightarrow\left(2^{2}\right)^{x-1} \times\left(\frac{1}{2}\right)^{3-2 x}=\left(\frac{1}{2^{3}}\right)^{x}$

$\Rightarrow\left(2^{2}\right)^{x-1} \times\left(2^{-1}\right)^{3-2 x}=\left(2^{-3}\right)^{x}$

$\Rightarrow 2^{2 x-2} \times 2^{2 x-3}=2^{-3 x}$

$\Rightarrow 2^{2 x-2+2 x-3}=2^{-3 x}$

$\Rightarrow 2^{4 x-5}=2^{-3 x}$

$\Rightarrow 4 x-5=-3 x$

$\Rightarrow 7 x=5$

$\Rightarrow x=\frac{5}{7}$

(vi)

$2^{3 x-7}=256$

$\Rightarrow 2^{3 x-7}=2^{8}$

$\Rightarrow 3 x-7=8$

$\Rightarrow 3 x=15$

$\Rightarrow x=5$

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