Solve the simultaneous equations:

$\displaystyle 64({4}^{y}) = {16}^{x} $

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

The answers are:

x = 1, y = –1

x = 2, y = 1

Thanks in advance.

Printable View

- May 1st 2011, 09:36 PMAngstyplanktonSimultaneous Exponential Equation Problem
Solve the simultaneous equations:

$\displaystyle 64({4}^{y}) = {16}^{x} $

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

The answers are:

x = 1, y = –1

x = 2, y = 1

Thanks in advance. - May 1st 2011, 10:58 PMsa-ri-ga-ma
You can write the first equation as

4^3*4^y = 4^2x.

From that you will get the value of y in-terms of x.

Substitute this value of y in the second equation. Solve the quadratics and find the values of e^x. From that get the required result.