Question

Solve the following equations:
(i) $3^{x+1}=27\times 3^4$
(ii) $4^{2x}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2$
(iii) $3^{x-1}\times 5^{2y-3}=225$
(iv) $8^{x+1}={16}^{y+2}\mathrm{and},{\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y}$
(v) $4^{x-1}\times {\left(0.5\right)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

(vi) $\sqrt{\frac{a}{b}}={\left(\frac{b}{a}\right)}^{1-2x}$, where a and b are distinct primes

Answer 1

(i)
$$\begin{gathered} 3^{x+1}=27\times 3^4 \\ \Rightarrow 3^{x+1}=3^3\times 3^4 \\ \Rightarrow 3^{x+1}=3^{3+4} \\ \Rightarrow 3^{x+1}=3^7 \\ \Rightarrow x+1=7 \\ \Rightarrow x=6 \end{gathered}$$

(ii)
$$\begin{gathered} 4^{2x}={\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2 \\ \Rightarrow 4^{2x}={\left(\sqrt{8}\right)}^2\mathrm{and}{\left(\sqrt[3]{16}\right)}^{-\frac{6}{y}}={\left(\sqrt{8}\right)}^2 \\ \Rightarrow 4^{2x}=\left(8^{\frac{1}{2}\times 2}\right)\mathrm{and}\left({16}^{\frac{1}{3}\times -\frac{6}{y}}\right)=\left(8^{\frac{1}{2}\times 2}\right) \\ \Rightarrow 4^{2x}=8\mathrm{and}\left({16}^{-\frac{2}{y}}\right)=8 \\ \Rightarrow 2^{4x}=2^3\mathrm{and}\left(2^{-\frac{8}{y}}\right)=2^3 \\ \Rightarrow 4x=3\mathrm{and}-\frac{8}{y}=3 \\ \Rightarrow x=\frac{3}{4}\mathrm{and}y=-\frac{8}{3} \end{gathered}$$

(iii)
$$\begin{gathered} 3^{x-1}\times 5^{2y-3}=225 \\ \Rightarrow 3^{x-1}\times 5^{2y-3}=3\times 3\times 5\times 5 \\ \Rightarrow 3^{x-1}\times 5^{2y-3}=3^2\times 5^2 \\ \Rightarrow x-1=2\mathrm{and}2y-3=2 \\ \Rightarrow x=3\mathrm{and}y=\frac{5}{2} \end{gathered}$$

(iv)
$$\begin{gathered} 8^{x+1}={16}^{y+2}\mathrm{and},{\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y} \\ \Rightarrow {\left(2^3\right)}^{x+1}={\left(2^4\right)}^{y+2}\mathrm{and}{\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{2^2}\right)}^{3y} \\ \Rightarrow {\left(2\right)}^{3x+3}={\left(2\right)}^{4y+8}\mathrm{and}{\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{2}\right)}^{6y} \\ \Rightarrow 3x+3=4y+8\mathrm{and}3+x=6y \end{gathered}$$
Now,
$3+x=6y\Rightarrow x=6y-3$
Putting x = 6y − 3 in $3x-4y=5$, we get
$$\begin{gathered} 3\left(6y-3\right)-4y=5 \\ \Rightarrow 18y-9-4y=5 \\ \Rightarrow 14y=14 \\ \Rightarrow y=1 \end{gathered}$$
Putting y = 1 in $x=6y-3$, we get
$x=6\times 1-3=3$

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

(vi)
$$\begin{gathered} \sqrt{\frac{a}{b}}={\left(\frac{b}{a}\right)}^{1-2x} \\ \Rightarrow {\left(\frac{a}{b}\right)}^{\frac{1}{2}}={\left(\frac{a}{b}\right)}^{-\left(1-2x\right)} \\ \Rightarrow \frac{1}{2}=-\left(1-2x\right) \\ \Rightarrow \frac{1}{2}=2x-1 \\ \Rightarrow \frac{3}{2}=2x \\ \Rightarrow x=\frac{3}{4} \end{gathered}$$