4^(x+2)=6^(-2x-3) I am completely clueless how to do this problem because neither my teacher nor my book explained it very well. Please HELP
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Make use of one of the rules of logs, $\displaystyle \log a^{k} = k\log a.$ It doesn't matter what base you choose, the rule works for any base. Take logs of your equation, apply the rule and then solve for x.
Originally Posted by paigesisco 4^(x+2)=6^(-2x-3) I am completely clueless how to do this problem because neither my teacher nor my book explained it very well. Write it as $\displaystyle 2^{2x+4}=(2^{-2x-3})(3^{-2x-3})$
Another way to do it, Remember that $\displaystyle 6= 4^{log_4 {6}}$ So we have $\displaystyle 4^{x+2}=4^{(log_4 {6})\cdot(-2x-3)}$ then $\displaystyle x+2= (log_4 {6})\cdot(-2x-3)$
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