3 years ago

A submarine at 21 feet below sea level suddenly moves up about 6 feet. At what depth is

Mathematics

1 answer:

Artist 52 [7]3 years ago

7 0

Answer:

The submarine is 15 feet below sea level.

Step-by-step explanation:

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In the first quadrant, you start at ( 6, 5 ) and move 3 units left .

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(3,5)

∧

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<img src="https://tex.z-dn.net/?f=%20log_%7B2%7D%283x%20%2B%204%29%20%20-%207%20log_%7B4%7D%7Bx%7D%5E%7B2%7D%20%20%2B%20%20log_%

olga2289 [7]

First of all, we need all logarithms to have the same base. So, we use the formula

$\log_a(b)=\dfrac{\log_c(b)}{\log_c(a)}$

To change the second term as follows:

$\log_4(x^2)=\dfrac{\log_2(x^2)}{\log_2(4)}=\dfrac{\log_2(x^2)}{2}$

Finally, using the property

$\log(a^b)=b\log(a)$

we have

$\dfrac{\log_2(x^2)}{2}=\log_2(x)$

So, the equation becomes

$\log_2(3x+4)-7\log_2(x)+\log_2(x)=2 \iff \log_2(3x+4)-6\log_2(x)=2$

We can now use the formula

$\log(a)-\log(b)=\log\left(\dfrac{a}{b}\right)$

to write the equation as

$\log_2(3x+4)-6\log_2(x)=2 \iff \log_2(3x+4)-\log_2(x^6)=2 \iff \log_2\left(\dfrac{3x+4}{x^6}\right)=2$

Now consider both sides as exponents of 2:

$\dfrac{3x+4}{x^6}=4 \iff 4x^6-3x-4=0$

This equation has no "nice" solution, so I guess the problem is as simplifies as it can be

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