Question

Find the values of x in each of the following:

(i) $2^{5x}÷2x=\sqrt[5]{2^{20}}$

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

(iii) ${\left(\frac{3}{5}\right)}^x{\left(\frac{5}{3}\right)}^{2x}=\frac{125}{27}$

(iv) $5^{x-2}\times 3^{2x-3}=135$

(v) $2^{x-7}\times 5^{x-4}=1250$

(vi) ${\left(\sqrt[3]{4}\right)}^{2x+\frac{1}{2}}=\frac{1}{32}$

(vii) $5^{2x+3}=1$

(viii) ${\left(13\right)}^{\sqrt{x}}=4^4-3^4-6$

(ix) ${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$

Answer 1

From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get

And,

Hence the value of x is

(vi) ${\left(\sqrt[3]{4}\right)}^{2x+\frac{1}{2}}=\frac{1}{32}$

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

On comparing we get,
$$\begin{gathered} \frac{4x+1}{3}=-5 \\ \Rightarrow 4x+1=-15 \\ \Rightarrow 4x=-16 \\ \Rightarrow x=-4 \end{gathered}$$

(vii) $5^{2x+3}=1$

$$\begin{gathered} 5^{2x+3}=5^0 \\ \Rightarrow 2x+3=0 \\ \Rightarrow x=\frac{-3}{2} \end{gathered}$$

(viii) ${\left(13\right)}^{\sqrt{x}}=4^4-3^4-6$

$$\begin{gathered} {\left(13\right)}^{\sqrt{x}}={\left(2^2\right)}^4-3^4-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=2^8-3^4-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=256-81-6 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}=169 \\ \Rightarrow {\left(13\right)}^{\sqrt{x}}={\left(13\right)}^2 \end{gathered}$$

On comparing we get,
$$\begin{gathered} \sqrt{x}=2 \\ \mathrm{on}\mathrm{squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \mathrm{x}=4 \end{gathered}$$

(ix) ${\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$
$$\begin{gathered} {\left(\sqrt{\frac{3}{5}}\right)}^{x+1}={\left(\frac{5}{3}\right)}^3 \\ \Rightarrow {\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{3}{5}\right)}^{-3} \end{gathered}$$

On comparing we get,

$$\begin{gathered} \frac{x+1}{2}=-3 \\ \Rightarrow x+1=-6 \\ \Rightarrow x=-7 \end{gathered}$$