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log_(7)64-log_(7)((8)/(3))+log_(7)2=log_(7)4p

$\log _{7} 64-\log _{7} \frac{8}{3}+\log _{7} 2=\log _{7} 4 p$

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Quotient property of logarithms

Full solution

Q.

\log _{7} 64-\log _{7} \frac{8}{3}+\log _{7} 2=\log _{7} 4 p

Rewrite log base $7$ : Use the change of base formula to rewrite $\log_{7}64$ as $\log_{7}2^6$ . $\newline$ $\log_{7}64 = \log_{7}(2^6) = 6 \cdot \log_{7}2$
Apply quotient rule: Apply the quotient rule of logarithms to $\log_{7}\left(\frac{8}{3}\right)$ . $\newline$ $\log_{7}\left(\frac{8}{3}\right) = \log_{7}8 - \log_{7}3$
Rewrite log base $7$ : Rewrite $\log_{7}8$ using the change of base formula. $\newline$ $\log_{7}8 = \log_{7}(2^3) = 3 \cdot \log_{7}2$
Substitute rewritten forms: Substitute the rewritten forms back into the original equation. $\log_{7}64 - \log_{7}\left(\frac{8}{3}\right) + \log_{7}2 = 6 \cdot \log_{7}2 - (3 \cdot \log_{7}2 - \log_{7}3) + \log_{7}2$
Simplify equation: Simplify the equation by distributing the negative sign and combining like terms. $\newline$ $6 \cdot \log_{7}2 - 3 \cdot \log_{7}2 + \log_{7}3 + \log_{7}2 = \log_{7}4p$
Combine terms with log base $7$ : Combine the terms with $\log_{7}2$ . $(6 - 3 + 1) \cdot \log_{7}2 + \log_{7}3 = \log_{7}4p$ $4 \cdot \log_{7}2 + \log_{7}3 = \log_{7}4p$
Use product rule: Use the product rule of logarithms to combine the left side of the equation. $\newline$ $\log_{7}(2^4) + \log_{7}3 = \log_{7}(16 \times 3) = \log_{7}48$ $\newline$ $\log_{7}48 = \log_{7}4p$
Solve for $p$ : Since the bases and the logs are the same, the arguments must be equal. $48 = 4p$
Solve for $p$ : Since the bases and the logs are the same, the arguments must be equal. $48 = 4p$ Divide both sides by $4$ to solve for $p$ . $p = \frac{48}{4}$ $p = 12$

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