Question

Find the value of x to make the following equation true.(3^2)^3 x 3^x = (3^7x3^3)/3^11Ax = 1Bx=-7сX = 8DX = -6

Answer 1

Given the expression:

$(3^2)^3\cdot3^x=\frac{3^7\cdot3^3}{3^{11}}$

notice that on the left side, if we simplify, we have the following:

$(3^2)^3\cdot3^x=3^6\cdot3^x=3^{x+6}$

and on the right side, we have:

$\frac{3^7\cdot3^3}{3^{11}}=\frac{3^{7+3}}{3^{11}}=\frac{3^{10}}{3^{11}}=3^{10-11}=3^{-1}$

then, if we equate both expression we have that:

$3^{x+6}=3^{-1}$

then, this must mean that the exponents are equal. This means the following:

$x+6=-1$

solving for x we get:

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

therefore, x = -7