Question

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

Answer 1

Answer:

(a) $\log_72 \approx 0.3562$

(b) $\log_7(\frac{147}{16} ) = 2$

Step-by-step 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:

$4\log_72 \approx 1.4248\\\log_72\approx \frac{1.4248}{4} \approx 0.3562$

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.

$\log_c\frac{a}{b} = \log_ca-\log_cb$

$\log_7\frac{147}{16} = \log_7147 -\log_716$

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:

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