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

Question

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

x3_drummerchick · Answer

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

solomonzelman · Answer

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

mathstudent55 · Answer

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  $

$\large x =    \dfrac{2 \log 3}  {\log 7 + 2 \log 3}$