text: ## Intermediate Algebra (6th Edition)
  
  $x=7$
  If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$. Therefore, $log_{x}49=2$ is equivalent to $x^{2}=49$. We can now solve for x. Take the square root of both sides. $(x^{2})^{\frac{1}{2}}=(49)^{\frac{1}{2}}$ $x=7$