Question

Given log_7⁡3≈0.5646 and log_7⁡16≈1.4248, evaluate the expressions. Please show work using given values.

Answer 1

(a)
`\log_72 \approx 0.3562`

(b)
`\log_7((147)/(16) ) = 2`

Explanation:

For the point a:

For this problem we use the information of
`\log_716 \approx 1.4248`

16 can be re-write as
`2^4`. Then you have:

`\log_72^4`

Using the power property of the logarithm we have:

`\log_ba^c = c\log_ba`

`\log_72^4 = 4\log_72`

And this expression is equal to:

`4\log_72 \approx 1.4248\\\log_72\approx (1.4248)/(4) \approx 0.3562\\`

For the point b:

For this problem we use the information of
`\log_716 \approx 1.4248` and
`\log_73 \approx 0.5646`

The first is use the quotient property in the given log.

`\log_c(a)/(b) = \log_ca-\log_cb`

`\log_7(147)/(16) = \log_7147 -\log_716`

We already know the value of
`\log_716` then we have to calculate the value of
`\log_7147`.

So we factorize the number 147 in
`3 \cdot 7 \cdot 7` (for factorize the number start with the smaller prime and check if the remainder is 0, it is then is a primer factor of the number).

Now re-write
`\log_7147` as
`\log_73 \cdot 7^2` and by the product property get:

`\log_bmn = \log_b m +\log_b n\\\log_73 \cdot 49 = \log_7 3 +\log_7 49 = 0.5646 + 2 = 2.5646`

Know replace each value in the subtraction and get the result:

`\log_7147 -\log_716 = 2.5646 - 0.5646 = 2`