i am having this problem and i couldn't prove that the left side equals the right side

i don't know what exactly that i am missing !!

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- Sep 6th 2013, 03:59 PMSHEvA9Equations::::
i am having this problem and i couldn't prove that the left side equals the right side

i don't know what exactly that i am missing !! - Sep 6th 2013, 04:12 PMHallsofIvyRe: Equations::::
Step one is to recognize that $\displaystyle 3^{-x}= \frac{1}{3^x}$ so that $\displaystyle \frac{1}{3^{-x}}= \frac{1}{\frac{1}{3^x}}= 3^x$. So $\displaystyle \frac{3^x}{5}- \frac{2}{3^{-x}}= \frac{3^x}{5}- 2(3^x)$. Now get common denominators by multiplying numerator and denominator of the last fraction by 5: $\displaystyle \frac{3^x}{5}- \frac{10(3^x)}{5}= \frac{-9(3^x)}{5}= -\frac{3^2(3^x)}{5}= -\frac{3^{2+ x}}{5}$.