## 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

1. anonymous

$7^{x/2}=3^{1-x}$

2. SolomonZelman

$$\large\color{black}{ \displaystyle 7^{x/2}=3^{1-x} }$$ $$\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3^{1-x} }$$ $$\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3(3)^{-x} }$$ $$\large\color{black}{ \displaystyle (\sqrt{7})^{x}=3\left(\frac{1}{3}\right)^{x} }$$ $$\large\color{black}{ \displaystyle 3^x(\sqrt{7})^{x}=3 }$$ $$\large\color{black}{ \displaystyle (3\sqrt{7})^{x}=3 }$$ I don't want to do it entirely for you, so from here take it on please.

3. mathstudent55

Why 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: $$\large x \log 7 = 2 \log 3 - 2x \log 3$$ $$\large x \log 7 + 2x \log 3= 2 \log 3$$ $$\large x( \log 7 + 2 \log 3)= 2 \log 3$$ $$\large x = \dfrac{2 \log 3} {\log 7 + 2 \log 3}$$