If $\log_7{5}=p$ and $\log_{10}{2} = q$ , solve $\log_7{2}$ in terms of $p$ and $q$

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Q. If

\log_7{5}=p

and

\log_{10}{2} = q

, solve

\log_7{2}

in terms of

p

and

q

Understand Given Information: Understand the given information and what is being asked. $\newline$ We are given that $\log_7{5}=p$ and $\log_{10}{2} = q$ . We need to express $\log_7{2}$ in terms of $p$ and $q$ .
Express Exponential Equations: Express $\log_7 5 = p$ as an exponential equation. $\newline$ $7^p = 5$
Relate Logarithms: Express $\log_{10}2=q$ as an exponential equation. $\newline$ $10^q = 2$
Break Down Expression: Express $\log 7^2$ in terms of $p$ . We need to find a way to relate $\log 7^2$ to $\log 7^5$ . We can do this by recognizing that $2$ is a factor of $5$ , specifically $5 = 2 \times (5/2)$ . So we can write: $\log 7^2 = \log 7(5 \times (2/5))$
Simplify Using Information: Use the properties of logarithms to break down $\log_7(5 \cdot (2/5))$ . $\log_7^2 = \log_7(5) + \log_7(2/5)$
Substitute Known Values: Simplify the expression using the given information. $\newline$ We know that $7^p = 5$ , so $\log_7(5) = p$ . Now we need to express $\log_7\left(\frac{2}{5}\right)$ in terms of $p$ and $q$ . $\newline$ $\log_7^2 = p + \log_7\left(\frac{2}{5}\right)$
Change Base: Break down $\log_7\left(\frac{2}{5}\right)$ further using the properties of logarithms. $\newline$ $\log_7\left(\frac{2}{5}\right) = \log_7(2) - \log_7(5)$
Simplify Further: Substitute the known values into the equation. $\newline$ We know that $\log_7(5) = p$ , so we can substitute that in. We also know that $10^q = 2$ , which means $\log_{10}(2) = q$ . We need to change the base from $10$ to $7$ for $\log_7(2)$ . $\newline$ $\log_7^2 = p + (\log_7(2) - p)$
Substitute Back: Change the base of $\log_{10}(2)$ to $\log_{7}(2)$ using the change of base formula. $\log_{7}(2) = \frac{\log_{10}(2)}{\log_{10}(7)}$ Since $\log_{10}(2) = q$ , we can substitute $q$ for $\log_{10}(2)$ . $\log_{7}(2) = \frac{q}{\log_{10}(7)}$
Final Simplification: Substitute $\log_7(2)$ back into the equation for $\log_7^2$ .
$\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)$
Final Simplification: Substitute $\log_7(2)$ back into the equation for $\log_7^2$ .
$\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)$ Simplify the expression.
Since $p - p = 0$ , we can remove them from the equation.
$\log_7^2 = \frac{q}{\log_{10}(7)}$
Final Simplification: Substitute $\log_7(2)$ back into the equation for $\log_7^2$ .
$\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)$ Simplify the expression.
Since $p - p = 0$ , we can remove them from the equation.
$\log_7^2 = \frac{q}{\log_{10}(7)}$ Recognize that $\log_{10}(7)$ is simply the reciprocal of $\log_7(10)$ .
$\log_7^2 = \frac{q}{(1 / \log_7(10))}$
Final Simplification: Substitute $\log_7(2)$ back into the equation for $\log_7^2$ .
$\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)$ Simplify the expression.
Since $p - p = 0$ , we can remove them from the equation.
$\log_7^2 = \frac{q}{\log_{10}(7)}$ Recognize that $\log_{10}(7)$ is simply the reciprocal of $\log_7(10)$ .
$\log_7^2 = \frac{q}{(1 / \log_7(10))}$ Simplify the expression by multiplying by the reciprocal.
$\log_7^2 = q \cdot \log_7(10)$
Final Simplification: Substitute $\log_7(2)$ back into the equation for $\log_7^2$ .
$\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)$ Simplify the expression.
Since $p - p = 0$ , we can remove them from the equation.
$\log_7^2 = \frac{q}{\log_{10}(7)}$ Recognize that $\log_{10}(7)$ is simply the reciprocal of $\log_7(10)$ .
$\log_7^2 = \frac{q}{1 / \log_7(10)}$ Simplify the expression by multiplying by the reciprocal.
$\log_7^2 = q \cdot \log_7(10)$ Recognize that $\log_7(10)$ is a constant and cannot be simplified further in terms of $\log_7^2$ $0$ and $\log_7^2$ $1$ .
Therefore, the final expression for $\log_7^2$ in terms of $\log_7^2$ $0$ and $\log_7^2$ $1$ is:
$\log_7^2 = q \cdot \log_7(10)$