What is the value of A when we rewrite ((5)/(6))^(x) as A^(-8x) ? Choose 1 answer: A) A=((5)/(6))^((1)/(8)) (B) A=((5)/(6))^(-(1)/(8)) (C) A=-8*(5)/(6) (D) A=(1)/(8)

Identify Base A: We want to express ((5)/(6))^x in the form of A^(-8x). To do this, we need to find a base A such that raising A to the power of -8x will give us the same result as ((5)/(6))^x.Equating the Expressions: To find A, we equate the two expressions: ((5)/(6))^x = A^(-8x)Setting Exponents Equal: Since the exponents must be the same for the bases to be equal, we can set the exponents equal to each other: x = -8x * log(A) to the base ((5)/(6))Isolating the Term: Divide both sides by x to isolate the term with A: 1 = -8 * log(A) to the base ((5)/(6))Solving for Logarithm: Now, divide both sides by -8 to solve for log(A) to the base ((5)/(6)): -1/8 = log(A) to the base ((5)/(6))Removing Logarithm: To remove the logarithm, we raise the base ((5)/(6)) to the power of both sides of the equation: ((5)/(6))^(-1/8) = AWe can now see that the correct expression for A is ((5)/(6))^(-1/8), which corresponds to answer choice (B).

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What is the value of
A when we rewrite
((5)/(6))^(x) as
A^(-8x) ?
Choose 1 answer:
A)
A=((5)/(6))^((1)/(8))
(B)

A=((5)/(6))^(-(1)/(8))
(C)
A=-8*(5)/(6)
(D)
A=(1)/(8)

What is the value of $A$ when we rewrite $\left(\frac{5}{6}\right)^{x}$ as $A^{-8x}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $A=\left(\frac{5}{6}\right)^{\frac{1}{8}}$ $\newline$ (B) $A=\left(\frac{5}{6}\right)^{-\frac{1}{8}}$ $\newline$ (C) $A=-8\cdot\frac{5}{6}$ $\newline$ (D) $A=\frac{1}{8}$

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Calculus

Find derivatives of using multiple formulae

Full solution

Q. What is the value of

A

when we rewrite

\left(\frac{5}{6}\right)^{x}

A^{-8x}

\newline

Choose

1

answer:

\newline

(A)

A=\left(\frac{5}{6}\right)^{\frac{1}{8}}

\newline

(B)

A=\left(\frac{5}{6}\right)^{-\frac{1}{8}}

\newline

(C)

A=-8\cdot\frac{5}{6}

\newline

(D)

A=\frac{1}{8}

Identify Base $A$ : We want to express $\left(\frac{5}{6}\right)^x$ in the form of $A^{-8x}$ . To do this, we need to find a base $A$ such that raising $A$ to the power of $-8x$ will give us the same result as $\left(\frac{5}{6}\right)^x$ .
Equating the Expressions: To find $A$ , we equate the two expressions: $\left(\frac{5}{6}\right)^x = A^{-8x}$
Setting Exponents Equal: Since the exponents must be the same for the bases to be equal, we can set the exponents equal to each other: $x = -8x \cdot \log_{\left(\frac{5}{6}\right)}(A)$
Isolating the Term: Divide both sides by $x$ to isolate the term with $A$ : $\newline$ $1 = -8 \cdot \log_{\left(\frac{5}{6}\right)}(A)$
Solving for Logarithm: Now, divide both sides by $-8$ to solve for $\log_{(\frac{5}{6})}(A)$ : $\newline$ $-\frac{1}{8} = \log_{(\frac{5}{6})}(A)$
Removing Logarithm: To remove the logarithm, we raise the base $\left(\frac{5}{6}\right)$ to the power of both sides of the equation: $\newline$ $\left(\frac{5}{6}\right)^{-\frac{1}{8}} = A$
Removing Logarithm: To remove the logarithm, we raise the base $\left(\frac{5}{6}\right)$ to the power of both sides of the equation: $\newline$ $\left(\frac{5}{6}\right)^{-\frac{1}{8}} = A$ We can now see that the correct expression for $A$ is $\left(\frac{5}{6}\right)^{-\frac{1}{8}}$ , which corresponds to answer choice $(B)$ .