Which of the following is equivalent to log_(2)(c)*log_(c)(2) ? Choose 1 answer: (A) log(c) (B) log(2) (c) 1 (D) -1

Recognize Relationship: Recognize the relationship between the two logarithms. The two logarithms log_(2)(c) and log_(c)(2) are inverses of each other because they have their bases and arguments swapped.Apply Change of Base: Apply the change of base formula to one of the logarithms. Change of base formula: log_(b)(a) = log_k(a) / log_k(b), where k is any positive number different from 1. Let's apply this to log_(c)(2): log_(c)(2) = log(2) / log(c).Substitute Change of Base: Substitute the change of base expression into the original equation. Now we have log_(2)(c) * (log(2) / log(c)).Simplify Expression: Simplify the expression. Notice that log_(2)(c) is equivalent to 1/log(c) in base 2. So we have (1 / log(c)) * (log(2) / log(c)).Recognize Expression Simplifies: Recognize that the expression simplifies to 1. The log(c) in the denominator of the first fraction and the log(c) in the numerator of the second fraction cancel each other out, leaving us with log(2) / log(2), which equals 1.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Which of the following is equivalent to
log_(2)(c)*log_(c)(2) ?
Choose 1 answer:
(A)
log(c)
(B)
log(2)
(c) 1
(D) -1

Which of the following is equivalent to $\log _{2}(c) \cdot \log _{c}(2)$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\log (c)$ $\newline$ (B) $\log (2)$ $\newline$ (C) $1$ $\newline$ (D) $-1$

View step-by-step help

Home

Math Problems

Algebra 2

Product property of logarithms

Full solution

Q. Which of the following is equivalent to

\log _{2}(c) \cdot \log _{c}(2)

\newline

Choose

1

answer:

\newline

(A)

\log (c)

\newline

(B)

\log (2)

\newline

(C)

1

\newline

(D)

-1

Recognize Relationship: Recognize the relationship between the two logarithms. $\newline$ The two logarithms $\log_{2}(c)$ and $\log_{c}(2)$ are inverses of each other because they have their bases and arguments swapped.
Apply Change of Base: Apply the change of base formula to one of the logarithms. $\newline$ Change of base formula: $\log_{b}(a) = \frac{\log_{k}(a)}{\log_{k}(b)}$ , where $k$ is any positive number different from $1$ . $\newline$ Let's apply this to $\log_{c}(2)$ : $\log_{c}(2) = \frac{\log(2)}{\log(c)}$ .
Substitute Change of Base: Substitute the change of base expression into the original equation. $\newline$ Now we have $\log_{2}(c) \cdot \left(\frac{\log(2)}{\log(c)}\right)$ .
Simplify Expression: Simplify the expression. $\newline$ Notice that $\log_{2}(c)$ is equivalent to $\frac{1}{\log(c)}$ in base $2$ . So we have $\left(\frac{1}{\log(c)}\right) \cdot \left(\frac{\log(2)}{\log(c)}\right)$ .
Recognize Expression Simplifies: Recognize that the expression simplifies to $1$ . $\newline$ The $\log(c)$ in the denominator of the first fraction and the $\log(c)$ in the numerator of the second fraction cancel each other out, leaving us with $\log(2) / \log(2)$ , which equals $1$ .