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Find $x$, in closed form, that satisfies: $$2^x=5^{x+2}$$

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$2^x=5^{x+2}$ $2^x=5^x\cdot 5^2$ $\left(\frac{2}{5}\right)^x=5^2$ $x\log\left(\frac{2}{5}\right)=2\log(5)$ $x=\frac{2\log(5)}{\log(2)-\log(5)}$
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greg1313 said: Solution: Spoiler $$\displaystyle 2^x \ = \ 5^{x+2}$$ $$\displaystyle log_2(2^x) \ = \ log_2(5^{x+2})$$ $$\displaystyle x \ = \ (x + 2)log_2(5)$$ $$\displaystyle x \ = \ xlog_2(5) \ + \ 2log_2(5)$$ $$\displaystyle x \ - \ xlog_2(5) \ = \ 2log_2(5)$$ $$\displaystyle x[1 \ - \ log_2(5)] \ = \ 2log_2(5)$$ $$\displaystyle x \ = \ \dfrac{2log_2(5)}{1 \ - \ log_2(5)}$$ My emphasis was to make sure that the final form of the answer would take up no more than two lines. Click to expand... That is how I did it (more or less).
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Solution: Spoiler $$\displaystyle 2^x \ = \ 5^{x+2}$$ $$\displaystyle log_2(2^x) \ = \ log_2(5^{x+2})$$ $$\displaystyle x \ = \ (x + 2)log_2(5)$$ $$\displaystyle x \ = \ xlog_2(5) \ + \ 2log_2(5)$$ $$\displaystyle x \ - \ xlog_2(5) \ = \ 2log_2(5)$$ $$\displaystyle x[1 \ - \ log_2(5)] \ = \ 2log_2(5)$$ $$\displaystyle x \ = \ \dfrac{2log_2(5)}{1 \ - \ log_2(5)}$$ My emphasis was to make sure that the final form of the answer would take up no more than two lines.
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Spoiler $$\displaystyle (\dfrac {2}{5})^x =25$$
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Debsta said: Spoiler Split, divide, log Click to expand... Spoiler: Spoiler I'm a lumberjack and I'm okay, I sleep all night and I work all day! -Dan
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Spoiler Split, divide, log
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