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# Algebra

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Solve the system of equations
y = \log_2 (2x)
y = log_4 (16 + x)

Jul 16, 2024

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First off, since both of them equal to y, then we can set the two equations to equal each other. We have

$$\log2 (2x) =\log4 (16 + x)$$

Applying the change of base thereom to the equation, we get

$$\log (2x) / \log2 = \log (16 + x) / \log 4 \\ \log (2x) / \log 2 = \log (16 + x) /\log 2^2 \\ \log (2x) / \log 2 = \log (16 + x) / [ 2 \log 2]$$

Now, we multiply through by log 2.

$$\log (2x) =\log (16 + x) / 2 \\ 2\log (2x) = \log (16 + x) \\ \log (2x)^2 = \log (16 + x)$$

Now, wait a second. Notice that both logs have the same base now.

This means there aren't any reprucussions we have to worry about. We simply just take the inside of the parenthesis.

This implies that

$$(2x)^2 = 16 + x \\ 4x^2 - x - 16 = 0$$

Taking the postiive value of x and using the quadratic equation, we get

The answer to the system of equations is $$x = [ 1 + \sqrt{257} ] / 8$$

Jul 16, 2024
edited by NotThatSmart  Jul 16, 2024