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.
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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.
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