Solve for the exact value of x. log_(4)(5x)+log_(4)(7)=3 Answer:

Combine logarithmic terms: Apply the product rule of logarithms to combine the two logarithmic terms. The product rule of logarithms states that log_b(m) + log_b(n) = log_b(m*n), where b is the base of the logarithms. So, log_4(5x) + log_4(7) becomes log_4(5x * 7). Calculation: log_4(5x * 7) = log_4(35x).Set equal and solve: Set the combined logarithm equal to 3 and solve for x. We have log_4(35x) = 3. To solve for x, we need to rewrite the equation in exponential form. The exponential form is b^y = x, where b is the base, y is the exponent, and x is the result. So, 4^3 = 35x. Calculation: 4^3 = 64, so 64 = 35x.Divide and find x: Divide both sides of the equation by 35 to solve for x. Calculation: x = 64 / 35.

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Solve for the exact value of
x.

log_(4)(5x)+log_(4)(7)=3
Answer:

Solve for the exact value of $x$ . $\newline$ $\log _{4}(5 x)+\log _{4}(7)=3$ $\newline$ Answer:

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Precalculus

Quotient property of logarithms

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Q. Solve for the exact value of

x

\newline

\log _{4}(5 x)+\log _{4}(7)=3

\newline

Answer:

Combine logarithmic terms: Apply the product rule of logarithms to combine the two logarithmic terms. $\newline$ The product rule of logarithms states that $\log_b(m) + \log_b(n) = \log_b(m\cdot n)$ , where $b$ is the base of the logarithms. $\newline$ So, $\log_4(5x) + \log_4(7)$ becomes $\log_4(5x \cdot 7)$ . $\newline$ Calculation: $\log_4(5x \cdot 7) = \log_4(35x)$ .
Set equal and solve: Set the combined logarithm equal to $3$ and solve for $x$ . We have $\log_4(35x) = 3$ . To solve for $x$ , we need to rewrite the equation in exponential form. The exponential form is $b^y = x$ , where $b$ is the base, $y$ is the exponent, and $x$ is the result. So, $4^3 = 35x$ . Calculation: $4^3 = 64$ , so $x$ $0$ .
Divide and find x: Divide both sides of the equation by $35$ to solve for $x$ . $\newline$ Calculation: $x = \frac{64}{35}$ .