What is the value of the expression shown below when a = 3 and b = 4 ? (a^2b^-3)(a^-3b^4) Select one: A. 4/3 B. 1 C. 3 D. 4
\(\bf (a^2b^{-3})(a^{-3}b^4)\qquad a={\color{red}{ 3}}\qquad b={\color{blue}{ 4}}\implies ({\color{red}{ 3}}^2{\color{blue}{ 4}}^{-3})({\color{red}{ 3}}^{-3}{\color{blue}{ 4}}^4) \\ \quad \\ \textit{keep in mind that } x^{-{\color{red} n}} = \cfrac{1}{x^{\color{red} n}}\)
so a?
hmm dunno... what did you get?
i got 242.25 and i know that isnt the answer
\(\bf (a^2b^{-3})(a^{-3}b^4)\qquad a={\color{red}{ 3}}\qquad b={\color{blue}{ 4}}\implies ({\color{red}{ 3}}^2{\color{blue}{ 4}}^{-3})({\color{red}{ 3}}^{-3}{\color{blue}{ 4}}^4) \\ \quad \\ \textit{keep in mind that } a^{-{\color{red} n}} = \cfrac{1}{a^{\color{red} n}}\qquad thus \\ \quad \\ ({\color{red}{ 3}}^2{\color{blue}{ 4}}^{-3})({\color{red}{ 3}}^{-3}{\color{blue}{ 4}}^4)\implies \Large \left({\color{red}{ 3}}^2\cdot \frac{1}{{\color{blue}{ 4}}^3}\right)\left(\frac{1}{{\color{red}{ 3}}^3}\cdot {\color{blue}{ 4}}^4\right)\)
so (6 x 1/64)(1/27 x 256)?
yeap
i got 4.5
@jdoe0001
hmm well
so its A?
yeap
have you covered the exponent rules yet?
a little
ok... notice using the exponent rules \(\bf ({\color{red}{ 3}}^2{\color{blue}{ 4}}^{-3})({\color{red}{ 3}}^{-3}{\color{blue}{ 4}}^4)\implies 3^2\cdot 3^{-3}\cdot 4^{-3}\cdot 4^4\implies 3^{2-3}\cdot 4^{-3+4} \\ \quad \\ 3^{-1}\cdot 4^1\implies \cfrac{1}{3^1}\cdot 4\implies \cfrac{4}{3}\)
oh ok
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