$\displaystyle 3^{x} = 2$ $\displaystyle 27^{-x} = ?$ The result is $\displaystyle \dfrac{1}{8}$ How do I get to the answer? Thanks!
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Apply these 2 rules: If a^p = b then p = log(b) / log(a) a^(-p) = 1 / a^p
Originally Posted by klik11 $\displaystyle 3^{x} = 2$ $\displaystyle 27^{-x} = ?$ The result is $\displaystyle \dfrac{1}{8}$ How do I get to the answer? Thanks! Slightly different way that avoids logs. $\displaystyle 27^{-x}=(3^3)^{-x}=3^{-3x}=(3^x)^{-3}=2^{-3}=\frac{1}{8}$
Thanks! I have one more exercise like this though, and I still can't solve it. $\displaystyle 3^{2x+8} = b^{8}$ $\displaystyle 9^{-x} = ?$ the result is $\displaystyle (\frac{b}{3})^{8}$
Originally Posted by klik11 Thanks! I have one more exercise like this though, and I still can't solve it. $\displaystyle 3^{2x+8} = b^{8}$ $\displaystyle 9^{-x} = ?$ the result is $\displaystyle (\frac{b}{3})^{8}$ $\displaystyle 3^{2x+8}=3^{2x}\cdot3^8=9^x\cdot3^8$ continue.. Actually I get $\displaystyle (\frac{3}{b})^{8}$
Yes my mistake, it's $\displaystyle (\frac{3}{b})^{8}$ Actually, I already figured that 9^x equals e^2x. I have no idea how to continue the exercise...
Originally Posted by klik11 Yes my mistake, it's $\displaystyle (\frac{3}{b})^{8}$ Actually, I already figured that 9^x equals e^2x. I have no idea how to continue the exercise... e^2x? Continuing from where I left off above $\displaystyle 9^x\cdot3^8=b^8$ $\displaystyle 9^x=(\frac{b}{3})^8$ $\displaystyle 9^{-x}=(\frac{3}{b})^8$
3^2x * Thanks!!!
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