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anonymous
 one year ago
how do i solve this? (will give medals!)
7^(x/2) = 3^(1 − x)
i think im supposed to use substitution in some way
anonymous
 one year ago
how do i solve this? (will give medals!) 7^(x/2) = 3^(1 − x) i think im supposed to use substitution in some way

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle 7^{x/2}=3^{1x} }\) \(\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3^{1x} }\) \(\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3(3)^{x} }\) \(\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3\left(\frac{1}{3}\right)^{x} }\) \(\large\color{black}{ \displaystyle 3^x(\sqrt{7})^{x}=3 }\) \(\large\color{black}{ \displaystyle (3\sqrt{7})^{x}=3 }\) I don't want to do it entirely for you, so from here take it on please.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.2Why not use logs? \(\Large 7^\frac{x}{2} = 3^{1 − x}\) Take the log of both sides: \(\large \log 7^\frac{x}{2} = \log 3^{1 − x}\) Use the exponent rule: \(\large \dfrac{x}{2} \log 7 = (1  x) \log 3\) Multiply both sides by 2: \(\large x \log 7 = (2  2x) \log 3\) Now work on isolating x: \(\large x \log 7 = 2 \log 3  2x \log 3\) \(\large x \log 7 + 2x \log 3= 2 \log 3 \) \(\large x( \log 7 + 2 \log 3)= 2 \log 3 \) \(\large x = \dfrac{2 \log 3} {\log 7 + 2 \log 3}\)
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