how do u solve 4^(3-x)=7?
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Originally Posted by Orlando how do u solve 4^(3-x)=7? in this case, by using logarithms $\displaystyle \log(4^{3-x}) = \log(7)$ now use the properties of logarithms to finish it
RULE: if a^x = b then x = log(b) / log(a)
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