I think it would be easy to find it because its 30 feet away and even if he didn't know his feet he could always leave rocks/pebbles to the treasure so he won't lose it.

6 0

3 years ago

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

2 years ago

A cone has the radius of 5 inches and height of 5 inches. What is the volume of the cone to the nearest tenth in3? Use 3.14

olasank [31]

Your answer would be 125.4 cause the number is too small to be 130 so yeah hope this helps

7 0

3 years ago

A woman starts out walking due north for 2 miles, and then she turns due west and continues for another 2 miles. How far from he

Mrac [35]

For this case we can model the problem as a rectangle triangle.
We have:
Length of the sides of the triangle.
We want to find:
Length of the hypotenuse.
Using the Pythagorean theorem we have:
$d ^ 2 = 2 ^ 2 + 2 ^ 2
$
$d = \sqrt{2 ^ 2 + 2 ^ 2} 

d = \sqrt{4 + 4} 

d = \sqrt{8} 

d = 2\sqrt{2} 
$
Answer:
She is:
$d = 2\sqrt{2}$
miles far from her starting point

4 0

3 years ago