### Instruction: Find $b$ if $\log_{b}343=-\frac{3}{2}$.### Response: I recognize 343 as a power of 7, namely $7^3$.
So I can rewrite the logarithm as $\log_{b}7^3=-\frac{3}{2}$.
Using the power rule of logarithms, I can move the exponent 3 down as a coefficient: $3\log_{b}7=-\frac{3}{2}$.
Dividing both sides by 3, I get $\log_{b}7=-\frac{1}{2}$.
This means that $b^{-\frac{1}{2}}=7$, or equivalently, $\frac{1}{\sqrt{b}}=7$.
To isolate $b$, I can take the reciprocal of both sides and then square them: $b=(\frac{1}{7})^2=\frac{1}{49}$.