Question

Given that
${3}^{3x + 1} \div {3}^{x} = 27$
find the value of x ​

Answer 1

Given:

• $\sf 3^{3x+1}\div 3^x=27$

To find:

• Value of x

Solution:

⇒ $\sf 3^{3x+1}\div 3^x=27$

⇒ $\sf 3^{3x+1}\div 3^x=3\times 3\times 3$

⇒ $\sf 3^{3x+1}\div 3^x=3^3$

⇒ $\sf 3^{3x+1-x}=3^3$

⇒ $\sf 3^{2x+1}=3^3$

Now base is same but exponents are different, so let's consider exponents.

⇒ $\sf 2x+1=3$

⇒ $\sf 2x=3-1$

⇒ $\sf 2x=2$

⇒ $\sf x=1$

So the required value of x is 1.

Additional Information:

$\boxed{\begin{array}{ l ll }\underline{\small{\maltese}\;{\textbf{\text{Law of Exponents :}}}}\\\\\star\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\star\:\:\sf{(a^m)^n = a^{mn}}\\\\\star\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\star\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\star\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\tfrac{1}{n}}\end{array}}$