write the expression

Question

Write the expression as a sum, difference, or multiple of logarithms. Assume that all variables represent positive real numbers.

 

1.  log₇ √6x

2. log₂ 16+ log₂ 4 - log₂ 2

3. 4√81 =3

Answer 1

1)

 

Convert the radical to a fraction exponent.

 

log₇(6x)^1/2

 

 

The exponents becomes the coefficient of the log.

 

(1/2)log₇(6x)

 

 

The product of the logs arguments is the sum of log.

 

(1/2)log₇(6) + (1/2)log₇(x)

 

 

 

2)

 

This is already expressed as a sum and difference.

 

 

3)

 

(81)^1/4 = 3

 

 

(1/4)log(81) = log(3)

 

 

Use product of the log's arguments on the left side.

 

(1/4)log(3² * 3²) = log(3)

 

 

Sum of the logs.

 

(1/4)log(3²) + (1/4)log(3²) = log(3)

 

 

Bring the exponent down as the coefficient of the logs.

 

(1/2)log(3) + (1/2)log(3) = log(3)

 

log(3) = log(3)

 

 

Answer 2

1) log₇√6x

1/2 log₇(6) + log₇(x)

 

If the x was meant to be under the square root as well then the correct answer is

1/2 (log₇(6) + log₇(x))

 

2) log₂16 + log₂4 + log₂2

Notice that 16 = 2⁴ and 4 = 2²

So we can rewrite the problem as

log₂2⁴ + log₂2² + log₂2

4 log₂2 + 2 log₂2 + log₂2 = 7 log₂2

 

3) Is there a typo here?