### Instruction: Find $b$ if $\log_{b}343=-\frac{3}{2}$.### Response: I want to rewrite this equation in exponential form, so I can get rid of the logarithm and solve for b.
To do that, I use the rule that $\log_b x = y$ is equivalent to $b^y = x$.
So, in this case, I have $\log_b 343 = -\frac{3}{2}$, which means $b^{-\frac{3}{2}} = 343$.
Now, I want to get rid of the negative exponent, so I can use the rule that $b^{-n} = \frac{1}{b^n}$.
This gives me $\frac{1}{b^{\frac{3}{2}}} = 343$.
Next, I want to isolate b, so I multiply both sides by $b^{\frac{3}{2}}$.
This gives me $1 = 343b^{\frac{3}{2}}$.
Then, I divide both sides by 343 to get $\frac{1}{343} = b^{\frac{3}{2}}$.
Now, I want to get rid of the fractional exponent, so I can use the rule that $b^{\frac{m}{n}} = \sqrt[n]{b^m}$.
This gives me $\frac{1}{343} = \sqrt[2]{b^3}$, or equivalently, $\frac{1}{343} = \left(b^3\right)^{\frac{1}{2}}$.
To undo the square root, I can square both sides of the equation.
This gives me $\frac{1}{117649} = b^3$.
Finally, to undo the cube, I can take the cube root of both sides of the equation.
This gives me $\sqrt[3]{\frac{1}{117649}} = b$.