Use the relationship x = a^y ? y = loga x to solve each of the following: log5 25 = x log3 x = 0 How would I go about doing those? I believe the 5 and 3 are the base.
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$\displaystyle log_{5}25=x$ $\displaystyle 5^x=25 (\because log_{a}b=c\iff a^c=b)$ $\displaystyle 5^x=5^2$ $\displaystyle \therefore x=2$ (comparing exponents) $\displaystyle log_{3}0=x$ $\displaystyle 3^x=0$ Which is impossible, so, no solution.
Originally Posted by Referos $\displaystyle log_{5}25=x$ $\displaystyle 5^x=25 (\because log_{a}b=c\iff a^c=b)$ $\displaystyle 5^x=5^2$ $\displaystyle \therefore x=2$ (comparing exponents) Exactly right. $\displaystyle log_{3}0=x$ $\displaystyle 3^x=0$ Which is impossible, so, no solution. No, the problem was $\displaystyle log_3 x= 0$ which is the same as $\displaystyle x= 3^0= 1$.
D'oh! Of course, you are correct; I misread the equation.
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