# Simultaneous Exponential Equation Problem

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• May 1st 2011, 10:36 PM
Angstyplankton
Simultaneous Exponential Equation Problem
Solve the simultaneous equations:

$64({4}^{y}) = {16}^{x}$
${3}^{y} = 4({3}^{x - 2}) - 1$

The answers are:

x = 1, y = –1
x = 2, y = 1

Thanks in advance.
• May 1st 2011, 11:58 PM
sa-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.