Question

Log_4^2(x)-log_4(x)-2=0

Answer 1

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Answer 2

To solve the equation $\log_4^2(x) - \log_4(x) - 2 = 0$, you can use the properties of logarithms.

First, notice that $\log_4^2(x)$ is the same as $(\log_4(x))^2$, so the equation becomes:

$(\log_4(x))^2 - \log_4(x) - 2 = 0$

Now, you can use a substitution to simplify this equation. Let $y = \log_4(x)$, so the equation becomes a quadratic equation in $y$:

$y^2 - y - 2 = 0$

Now, you can factor this quadratic equation:

$(y - 2)(y + 1) = 0$

Set each factor equal to zero:

1. $y - 2 = 0$
   $y = 2$

2. y + 1 = 0) \(y = -1

Now, remember that $y = \log_4(x)$. So, you have:

1. $\log_4(x) = 2$

2. $\log_4(x) = -1$

Now, you can solve for $x$ in each of these equations:

1. $\log_4(x) = 2$

Use the definition of a logarithm to rewrite this as an exponential equation:

$4^2 = x$

$16 = x$

2. $\log_4(x) = -1$

Again, use the definition of a logarithm:

$4^{-1} = x$

$\frac{1}{4} = x$

So, the solutions to the equation are $x = 16$ and $x = \frac{1}{4}$.