# Math Help - elimination

1. ## elimination

x = e^(2t), y=e^(3t)+8

2. Hello, pourcarlos!

Eliminate the parameter: . $\begin{array}{ccc}x \:=\:e^{2t} & [1] \\ y\:=\:e^{3t}+8 & [2]\end{array}$

From [1]: . $x \:=\:e^{2t} \quad\Rightarrow\quad 2t \:=\:\ln x \quad\Rightarrow\quad t \:=\:\tfrac{1}{2}\ln x$

Substitute into [2]: . $y \;=\;e^{3(\frac{1}{2}\ln x)}+8 \;=\;e^{(\frac{3}{2}\ln x)} + 8 \;=\;e^{(\ln x^{\frac{3}{2}})} + 8$

Therefore: . $y \;=\;x^{\frac{3}{2}} + 8$

3. oh great, thank you

4. Originally Posted by pourcarlos
x = e^(2t), y=e^(3t)+8
An alternative to the perfectly correct solution provide by Soroban:

$x^3 = e^{6t}$ and $(y - 8)^2 = e^{6t}$. Therefore ....

How much of the curve is required will obviously depend on the values of t that are allowed. If $t \geq 0$ then the curve starts at (1, 9) ....