How can you express 6log_2(m) + log_2(x)/3 as a single logarithm?

How can you express 6log_2(m) + log_2(x)/3 as a single logarithm?

• A. log_2(m^6 * cbrt(x))
• B. log_2(m^6 * x^(1/3))
• C. log_2(m^6 + x^(1/3))
• D. log_2(m^3 * x^2)

Answer: (A)

To combine the expression \( 6\log_2(m) + \frac{\log_2(x)}{3} \) into a single logarithm, we can use the properties of logarithms. The property \( a\log_b(c) = \log_b(c^a) \) allows us to move coefficients in front of logs into the exponent of their arguments. This means: 1. The term \( 6\log_2(m) \) can be expressed as \( \log_2(m^6) \). 2. The term \( \frac{\log_2(x)}{3} \) can be rewritten as \( \log_2(x^{1/3}) \) because \( \frac{1}{3} \) is the same as raising \( x \) to the power of \( \frac{1}{3} \). Now, we have converted the original expression into: \[ \log_2(m^6) + \log_2(x^{1/3}) \] Using the logarithmic property \( \log_b(a) + \log_b(c) = \log_b(a \cdot c) \), we can combine these two logarithms:

To combine the expression ( 6\log_2(m) + \frac{\log_2(x)}{3} ) into a single logarithm, we can use the properties of logarithms.

The property ( a\log_b(c) = \log_b(c^a) ) allows us to move coefficients in front of logs into the exponent of their arguments. This means:

1. The term ( 6\log_2(m) ) can be expressed as ( \log_2(m^6) ).

2. The term ( \frac{\log_2(x)}{3} ) can be rewritten as ( \log_2(x^{1/3}) ) because ( \frac{1}{3} ) is the same as raising ( x ) to the power of ( \frac{1}{3} ).

Now, we have converted the original expression into:

[

\log_2(m^6) + \log_2(x^{1/3})

]

Using the logarithmic property ( \log_b(a) + \log_b(c) = \log_b(a \cdot c) ), we can combine these two logarithms: