# Thread: logarithms equation

1. ## logarithms equation

The curves $\displaystyle y = \frac{1}{3}^x$ and $\displaystyle y = 2(3^x)$ intersect at the point P.
(c) Find the x-coordinate of P to 2 decimal places and show that the y-coordinate
of P is $\displaystyle \sqrt{2}$

$\displaystyle \frac{1}{3}^x = 2 (3^x)$

$\displaystyle x = \frac{lg\frac{1}{2}}{2lg3} = -0.32$

How do I show the corresponding 'y' value = $\displaystyle \sqrt{2}$ ?

$\displaystyle y = 2(3^{-0.32})$ this equals the exact value of root 2, but I need to show it using Algebra?

can anyone show me?

Thanks!

2. Hello, Tweety!

The curves .$\displaystyle \begin{Bmatrix}y &=&2\cdot3^x & {\color{blue}[1]} \\ y &=& \left(\tfrac{1}{3}\right)^x & {\color{blue}[2]} \end{Bmatrix}$ . intersect at point $\displaystyle P.$

(c) Find the $\displaystyle x$-coordinate of $\displaystyle P$ to 2 decimal places
. . .and show that the $\displaystyle y$-coordinate of $\displaystyle P$ is $\displaystyle \sqrt{2}.$

Equate [1] and [2]: .$\displaystyle 2\cdot3^x \:=\:\frac{1}{3^x}$

Multiply by $\displaystyle 3^x\!:\quad 2\cdot3^{2x} \:=\:1 \quad\Rightarrow\quad 3^{2x} \:=\:\frac{1}{2}$ .[3]

Take logs: .$\displaystyle \ln(3^{2x}) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln(2^{\text{-}1}) \quad\Rightarrow\quad 2x\ln(3)\:=\:-\ln(2)$

. . $\displaystyle x \;=\;-\frac{\ln(2)}{2\ln(3)} \;=\;-0.315464877 \;\approx\;\boxed{-0.32}$

At [3], we have: .$\displaystyle 3^{2x} \:=\:\frac{1}{2}$

Take logs (base 3): .$\displaystyle \log_3(3^{2x}) \:=\:\log_3\left(\frac{1}{2}\right) \:=\:\log_3(2^{-1})$

$\displaystyle \text{Then we have: }\;2x\underbrace{\log_3(3)}_{\text{This is 1}} \;=\;-\log_3(2) \quad\Rightarrow\quad 2x \:=\:-\log_3(2)$

. . $\displaystyle x \:=\:-\tfrac{1}{2}\log_3(2) \:=\:\log_3\!\left(2^{-\frac{1}{2}}\right) \:=\:\log_3\!\left(\frac{1}{2^{\frac{1}{2}}}\right ) \quad\Rightarrow\quad x\:=\:\log_3\left(\frac{1}{\sqrt{2}}\right)$

Substitute into [1]: .$\displaystyle y \;=\;2\cdot3^{\log_3\left(\frac{1}{\sqrt{2}}\right )} \;=\;2\cdot\frac{1}{\sqrt{2}} \quad\Rightarrow\quad y \;=\;\sqrt{2}$