Let 0 < x < 1. If y=log_(7)((1)/(x)), then (1)/(y)=

Express y in terms of x: Let's start by expressing y in terms of x using the given logarithmic equation y = log_7(1/x).Use logarithmic function property: We know that the logarithmic function log_b(a) = c means that b^c = a. So, in our case, 7^y = 1/x.Find reciprocal of 7^y: Now, we want to find the value of 1/y. To do this, we can take the reciprocal of both sides of the equation 7^y = 1/x to get 7^(-y) = x.Determine -y in terms of x: Since 7^(-y) is the reciprocal of 7^y, and we have 7^(-y) = x, it follows that -y is the logarithm base 7 of x, or -y = log_7(x).Calculate 1/y: Therefore, 1/y is the negative reciprocal of log_7(x), which means 1/y = -1/(log_7(x)).Consider original expression: However, we need to be careful here. The original question asks for 1/y in terms of the original expression y=log_7(1/x). We have found that 1/y = -1/(log_7(x)), but we need to express it in terms of the original variable, which is 1/x, not x.Analyze log_7(1/x): Since x is between 0 and 1, 1/x is greater than 1. Therefore, log_7(1/x) is negative because 7 raised to any positive power will be greater than 1, and we are taking the log of a number less than 1.Determine sign of 1/y: Given that y = log_7(1/x) and y is negative, 1/y is the negative reciprocal of a negative number, which is positive. So, 1/y = -1/(log_7(1/x)).Express 1/y in terms of y: Finally, we can simplify this to 1/y = -1/(-y) since y = log_7(1/x).Simplify the expression: Simplifying further, we get 1/y = 1/y, which means that 1/y is simply the reciprocal of y.

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Q. Let

0<x<1

. If

y=\log _{7} \frac{1}{x}

, then

\frac{1}{y}=

Express $y$ in terms of $x$ : Let's start by expressing $y$ in terms of $x$ using the given logarithmic equation $y = \log_7(\frac{1}{x})$ .
Use logarithmic function property: We know that the logarithmic function $\log_b(a) = c$ means that $b^c = a$ . So, in our case, $7^y = \frac{1}{x}.$
Find reciprocal of $7^y$ : Now, we want to find the value of $\frac{1}{y}$ . To do this, we can take the reciprocal of both sides of the equation $7^y = \frac{1}{x}$ to get $7^{-y} = x$ .
Determine $-y$ in terms of $x$ : Since $7^{(-y)}$ is the reciprocal of $7^y$ , and we have $7^{(-y)} = x$ , it follows that $-y$ is the logarithm base $7$ of $x$ , or $-y = \log_7(x)$ .
Calculate $1/y$ : Therefore, $1/y$ is the negative reciprocal of $\log_7(x)$ , which means $1/y = -1/(\log_7(x))$ .
Consider original expression: However, we need to be careful here. The original question asks for $\frac{1}{y}$ in terms of the original expression $y=\log_7(\frac{1}{x})$ . We have found that $\frac{1}{y} = -\frac{1}{\log_7(x)}$ , but we need to express it in terms of the original variable, which is $\frac{1}{x}$ , not $x$ .
Analyze $\log_7(\frac{1}{x})$ : Since $x$ is between $0$ and $1$ , $\frac{1}{x}$ is greater than $1$ . Therefore, $\log_7(\frac{1}{x})$ is negative because $7$ raised to any positive power will be greater than $1$ , and we are taking the log of a number less than $1$ .
Determine sign of $1/y$ : Given that $y = \log_7(1/x)$ and $y$ is negative, $1/y$ is the negative reciprocal of a negative number, which is positive. So, $1/y = -1/(\log_7(1/x))$ .
Express $1/y$ in terms of $y$ : Finally, we can simplify this to $1/y = -1/(-y)$ since $y = \log_7(1/x)$ .
Simplify the expression: Simplifying further, we get $\frac{1}{y} = \frac{1}{y}$ , which means that $\frac{1}{y}$ is simply the reciprocal of $y$ .