Question

Solve the following equations for x:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer 1

(i)
$$\begin{gathered} 7^{2x+3}=1 \\ \Rightarrow 7^{2x+3}=7^0 \\ \Rightarrow 2x+3=0 \\ \Rightarrow 2x=-3 \\ \Rightarrow x=-\frac{3}{2} \end{gathered}$$

(ii)
$$\begin{gathered} 2^{x+1}=4^{x-3} \\ \Rightarrow 2^{x+1}={\left(2^2\right)}^{x-3} \\ \Rightarrow 2^{x+1}=\left(2^{2x-6}\right) \\ \Rightarrow x+1=2x-6 \\ \Rightarrow x=7 \end{gathered}$$

(iii)
$$\begin{gathered} 2^{5x+3}=8^{x+3} \\ \Rightarrow 2^{5x+3}={\left(2^3\right)}^{x+3} \\ \Rightarrow 2^{5x+3}=2^{3x+9} \\ \Rightarrow 5x+3=3x+9 \\ \Rightarrow 2x=6 \\ \Rightarrow x=3 \end{gathered}$$

(iv)
$$\begin{gathered} 4^{2x}=\frac{1}{32} \\ \Rightarrow {\left(2^2\right)}^{2x}=\frac{1}{2^5} \\ \Rightarrow 2^{4x}\times 2^5=1 \\ \Rightarrow 2^{4x+5}=2^0 \\ \Rightarrow 4x+5=0 \\ \Rightarrow x=-\frac{5}{4} \end{gathered}$$

(v)
$$\begin{gathered} 4^{x-1}\times {\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^x \\ \Rightarrow {\left(2^2\right)}^{x-1}\times {\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{2^3}\right)}^x \\ \Rightarrow {\left(2^2\right)}^{x-1}\times {\left(2^{-1}\right)}^{3-2x}={\left(2^{-3}\right)}^x \\ \Rightarrow 2^{2x-2}\times 2^{2x-3}=2^{-3x} \\ \Rightarrow 2^{2x-2+2x-3}=2^{-3x} \\ \Rightarrow 2^{4x-5}=2^{-3x} \\ \Rightarrow 4x-5=-3x \\ \Rightarrow 7x=5 \\ \Rightarrow x=\frac{5}{7} \end{gathered}$$

(vi)
$$\begin{gathered} 2^{3x-7}=256 \\ \Rightarrow 2^{3x-7}=2^8 \\ \Rightarrow 3x-7=8 \\ \Rightarrow 3x=15 \\ \Rightarrow x=5 \end{gathered}$$