Question

log₇(x) - 1 - log₇(x)² + log₇(x²) divided by log₇(x)² + log₇(x²)

Explanation:

Let \( t = \log_{7}(x) \). First, simplify the numerator and the denominator using logarithm properties. Recall that \( \log_{a}(b^{c}) = c\log_{a}(b) \), so \( \log_{7}(x^{2}) = 2\log_{7}(x)=2t \).

Step 1: Substitute \( \log_{7}(x^{2}) \)

Substitute \( \log_{7}(x^{2}) = 2t \) into the expression. The numerator becomes \( t - 1 - t^{2}+ 2t \), and the denominator becomes \( t^{2}+ 2t \).

Step 2: Simplify the numerator

Simplify the numerator: \( t - 1 - t^{2}+ 2t=-t^{2}+3t - 1 \). The denominator is \( t^{2}+ 2t=t(t + 2) \).

Wait, maybe we can also simplify before substitution. Let's re - examine the original expression:

Original expression: \( \frac{\log_{7}(x)-1-\log_{7}(x)^{2}+\log_{7}(x^{2})}{\log_{7}(x)^{2}+\log_{7}(x^{2})} \)

Using \( \log_{7}(x^{2}) = 2\log_{7}(x) \), let \( y=\log_{7}(x) \), then the expression is:

\( \frac{y - 1 - y^{2}+2y}{y^{2}+2y}=\frac{-y^{2}+3y - 1}{y^{2}+2y} \)

We can also factor numerator and denominator (if possible). The denominator factors as \( y(y + 2) \). The numerator \( -y^{2}+3y - 1=-(y^{2}-3y + 1) \). The quadratic \( y^{2}-3y + 1 \) has roots \( y=\frac{3\pm\sqrt{9 - 4}}{2}=\frac{3\pm\sqrt{5}}{2} \), but maybe we made a mistake in the initial approach. Wait, maybe the problem is to simplify the expression. Let's try to combine like terms again.

Wait, \( \log_{7}(x)-1-\log_{7}(x)^{2}+\log_{7}(x^{2})=\log_{7}(x)+\log_{7}(x^{2})-\log_{7}(x)^{2}-1 \)

Using \( \log_{a}(M)+\log_{a}(N)=\log_{a}(MN) \), \( \log_{7}(x)+\log_{7}(x^{2})=\log_{7}(x\cdot x^{2})=\log_{7}(x^{3}) = 3\log_{7}(x) \)

So the numerator is \( 3\log_{7}(x)-\log_{7}(x)^{2}-1 \), and the denominator is \( \log_{7}(x)^{2}+2\log_{7}(x) \)

Let \( t = \log_{7}(x) \), then numerator: \( -t^{2}+3t - 1 \), denominator: \( t^{2}+2t \)

We can write the fraction as \( \frac{-t^{2}+3t - 1}{t^{2}+2t}=\frac{-(t^{2}-3t + 1)}{t(t + 2)} \)

If we want to further simplify, we can perform polynomial long - division on the numerator and the denominator. Divide \( -t^{2}+3t - 1 \) by \( t^{2}+2t \)

\( \frac{-t^{2}+3t - 1}{t^{2}+2t}=\frac{-(t^{2}+2t)+5t - 1}{t^{2}+2t}=-1+\frac{5t - 1}{t^{2}+2t} \)

Or factor the denominator \( t^{2}+2t=t(t + 2) \) and the numerator \( 5t - 1 \) can't be factored with the denominator.

Wait, maybe there was a misinterpretation of the original problem. If the problem is to simplify the expression, the simplified form (after substitution and combining like terms) is \( \frac{-(\log_{7}(x))^{2}+3\log_{7}(x)-1}{(\log_{7}(x))^{2}+2\log_{7}(x)} \) or by dividing numerator and denominator by \( \log_{7}(x) \) (assuming \( \log_{7}(x)
eq0 \)):

\( \frac{-\log_{7}(x)+3-\frac{1}{\log_{7}(x)}}{\log_{7}(x)+2} \) (for \( \log_{7}(x)
eq0 \))

But maybe the problem has a typo or we missed something. Alternatively, if we consider that maybe the original expression is supposed to be simplified by canceling terms, but with the given expression, the simplified form (after substituting \( t = \log_{7}(x) \) and \( \log_{7}(x^{2}) = 2t \)) is \( \frac{-t^{2}+3t - 1}{t^{2}+2t} \) where \( t=\log_{7}(x) \)

Answer:

\( \frac{-(\log_{7}(x))^{2}+3\log_{7}(x)-1}{(\log_{7}(x))^{2}+2\log_{7}(x)} \) (or equivalent forms after further manipulation)