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

Answer:

(a) [tex]\log_72 \approx 0.3562[/tex]

(b) [tex]\log_7(\frac{147}{16} ) = 2[/tex]

Step-by-step explanation:

For the point a:

For this problem we use the information of [tex]\log_716 \approx 1.4248[/tex]

16 can be re-write as [tex]2^4[/tex]. Then you have:

[tex]\log_72^4[/tex]

Using the power property of the logarithm we have:

[tex]\log_ba^c = c\log_ba[/tex]

[tex]\log_72^4 = 4\log_72[/tex]

And this expression is equal to:

[tex]4\log_72 \approx 1.4248\\\log_72\approx \frac{1.4248}{4} \approx 0.3562\\[/tex]

For the point b:

For this problem we use the information of [tex]\log_716 \approx 1.4248[/tex] and [tex]\log_73 \approx 0.5646[/tex]

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

[tex]\log_c\frac{a}{b} = \log_ca-\log_cb[/tex]

[tex]\log_7\frac{147}{16} = \log_7147 -\log_716[/tex]

We already know the value of [tex]\log_716[/tex] then we have to calculate the value of [tex]\log_7147[/tex].

So we factorize the number 147 in [tex]3 \cdot 7 \cdot 7[/tex] (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 [tex]\log_7147[/tex] as [tex]\log_73 \cdot 7^2[/tex] and by the product property get:

[tex]\log_bmn = \log_b m +\log_b n\\\log_73 \cdot 49 = \log_7 3 +\log_7 49 = 0.5646 + 2 = 2.5646[/tex]

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

[tex]\log_7147 -\log_716 = 2.5646 - 0.5646 = 2[/tex]

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