• Dec 6th 2009, 11:48 AM
mbasit
Need help in solving this simultaneous equations:
from add maths book by Ho Soo Thong
(64) (4^y) = 16^x -------eq 1 (I have already reduced it to y=2x-3)

3^y = 4[3^(x-2)] - 1 --------eq 2

Have tried many different ways but to no avail. Kindly help
• Dec 6th 2009, 12:01 PM
earboth
Quote:

Originally Posted by mbasit
Need help in solving this simultaneous equations:
from add maths book by Ho Soo Thong
(64) (4^y) = 16^x -------eq 1 (I have already reduced it to y=2x-3)

3^y = 4[3^(x-2)] - 1 --------eq 2

Have tried many different ways but to no avail. Kindly help

Substitute y by 2x - 3 in the 2nd equation:

$\displaystyle 3^y=4 \cdot 3^{x-2} - 1$

$\displaystyle 3^{2x-3}=4 \cdot 3^{x-2} - 1$

$\displaystyle \dfrac{3^{2x}}{27}=\dfrac49 \cdot 3^x - 1$ Multiply by 27

$\displaystyle 3^{2x}-12 \cdot 3^x+27=0$

This is a quadratic equation in $\displaystyle 3^x$. Substitute $\displaystyle z = 3^x$

$\displaystyle z^2-12z+27=0~\implies~z = 3~\vee~z = 9$

Resubstitute to get x = 2 or x = 1