If \(3^{x-1}+3^{x+1}=90\), then \(x=?\)

Given:  \(3^{x-1}+3^{x+1}=90\)

Step 1: Simplify the expression in L.H.S \(\Rightarrow 3^x \cdot 3^{-1}+3^x \cdot 3=90\)      [ \(∵ a^{m+n}=a^m \cdot a^n \)] By taking \( 3^x\) common, we get \(3^x(3^{-1}+ 3)=90\) \(\Rightarrow 3^x (\frac{1}{3}+3)=90\)          [ \(∵ a^{-1}=\frac{1}{a} \)] \(\Rightarrow 3^x (\frac{1+9}{3})=90\) \(\Rightarrow 3^x (\frac{10}{3})=90\) Step 2: By using laws of exponents Solve the equation for \(x\). \(3^x=90 \times \frac{3}{10}\) \(\Rightarrow 3^x=27\) \(\Rightarrow 3^x=3^3\) \(\Rightarrow x=3\)                  [ \(∵ a^m=a^n \iff m=n \)] Therefore, if \(3^{x-1}+3^{x+1}=90\) then \(x=3\).