### Instruction: Find $b$ if $\log_{b}343=-\frac{3}{2}$.### Response: To solve for $b$, I need to rewrite the logarithmic equation in exponential form, using the definition of logarithm: $\log_b x=y$ means $b^y=x$.
So, applying this definition to the given equation, I get $b^{-\frac{3}{2}}=343$.
Now, I need to find a way to simplify both sides of the equation, preferably using the same base.
I notice that 343 is a perfect cube: $343=7^3$.
So, I can rewrite the right-hand side as $7^3$.
On the left-hand side, I can use the property of exponents that says $b^{-\frac{3}{2}}=\frac{1}{b^{\frac{3}{2}}}$.
And then, I can use another property of exponents that says $b^{\frac{3}{2}}=\sqrt{b^3}$.
So, I can rewrite the left-hand side as $\frac{1}{\sqrt{b^3}}$.
Now, I have the equation $\frac{1}{\sqrt{b^3}}=7^3$.
To solve for $b$, I can cross-multiply and get $\sqrt{b^3}=7^{-3}$.
Then, I can square both sides and get $b^3=7^{-6}$.
And finally, I can take the cube root of both sides and get $b=7^{-2}$.