#### Answer

$.7^{3}=.343$

#### Work Step by Step

We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number).
Therefore, $log_{.7}.343=3$ can be written as the exponential equation $.7^{3}=.343$.