2^(x+4)-2^(x+1)=3

 

Can tackle it like this:

 

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

 

$$2^x2^4-2^x2^1=3$$

 

$$2^x(2^4-2^1)=3$$

 

$$2^x(16-2)=3$$

 

$$2^x\times14=3$$

 

$$2^x=\frac{3}{14}$$

 

$$\ln{2^x}=\ln{\frac{3}{14}}$$

 

$$x\times \ln2=\ln3-\ln14$$

 

$$x=\frac{\ln3-\ln14}{\ln2}$$

 

$${\mathtt{x}} = {\frac{\left({ln}{\left({\mathtt{3}}\right)}{\mathtt{\,-\,}}{ln}{\left({\mathtt{14}}\right)}\right)}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{x}} = -{\mathtt{2.222\: \!392\: \!421\: \!336\: \!447\: \!9}}$$

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$${{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)} = {\mathtt{3}} \Rightarrow {\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({ln}{\left({\mathtt{2}}\right)}{\mathtt{\,-\,}}{ln}{\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}\right)}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{x}} = -{\mathtt{2.222\: \!392\: \!421\: \!336\: \!448\: \!1}}$$

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