We can expand the logarithm of a product as a sum of logarithms:
![\log_dabc=\log_da+\log_db+\log_dc](https://tex.z-dn.net/?f=%5Clog_dabc%3D%5Clog_da%2B%5Clog_db%2B%5Clog_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}](https://tex.z-dn.net/?f=%5Clog_xy%3D%5Cdfrac%7B%5Cln%20y%7D%7B%5Cln%20x%7D%3D%5Cdfrac1%7B%5Cfrac%7B%5Cln%20x%7D%7B%5Cln%20y%7D%7D%3D%5Cdfrac1%7B%5Clog_yx%7D)
This immediately tells us that
![\log_dc=\dfrac1{\log_cd}=\dfrac12](https://tex.z-dn.net/?f=%5Clog_dc%3D%5Cdfrac1%7B%5Clog_cd%7D%3D%5Cdfrac12)
Notice that none of
can be equal to 1. This is because
![\log_1x=y\implies1^{\log_1x}=1^y\implies x=1](https://tex.z-dn.net/?f=%5Clog_1x%3Dy%5Cimplies1%5E%7B%5Clog_1x%7D%3D1%5Ey%5Cimplies%20x%3D1)
for any choice of
. 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}](https://tex.z-dn.net/?f=%5Clog_db%3D%5Cdfrac%7B%5Cln%20b%7D%7B%5Cln%20d%7D%3D%5Cdfrac%7B%5Cfrac%7B%5Cln%20b%7D%7B%5Cln%20c%7D%7D%7B%5Cfrac%7B%5Cln%20d%7D%7B%5Cln%20c%7D%7D%3D%5Cdfrac%7B%5Clog_cb%7D%7B%5Clog_cd%7D%3D%5Cdfrac1%7B%5Clog_bc%5Clog_cd%7D)
so that
![\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23](https://tex.z-dn.net/?f=%5Clog_db%3D%5Cdfrac1%7B-%5Cfrac34%5Ccdot2%7D%3D-%5Cdfrac23)
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}](https://tex.z-dn.net/?f=%5Clog_da%3D%5Cdfrac%7B%5Cln%20a%7D%7B%5Cln%20d%7D%3D%5Cdfrac%7B%5Cfrac%7B%5Cln%20a%7D%7B%5Cln%20b%7D%7D%7B%5Cfrac%7B%5Cln%20d%7D%7B%5Cln%20b%7D%7D%3D%5Cdfrac%7B%5Clog_ba%7D%7B%5Clog_bd%7D%3D%5Cdfrac%7B%5Clog_db%7D%7B%5Clog_ab%7D)
so that
![\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34](https://tex.z-dn.net/?f=%5Clog_da%3D%5Cdfrac%7B-%5Cfrac23%7D%7B%5Cfrac89%7D%3D-%5Cdfrac34)
So we end up with
![\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}](https://tex.z-dn.net/?f=%5Clog_dabc%3D-%5Cdfrac34-%5Cdfrac23%2B%5Cdfrac12%3D-%5Cdfrac%7B11%7D%7B12%7D)
###
Another way to do this:
![\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}](https://tex.z-dn.net/?f=%5Clog_ab%3D%5Cdfrac89%5Cimplies%20a%5E%7B8%2F9%7D%3Db%5Cimplies%20a%3Db%5E%7B9%2F8%7D)
![\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}](https://tex.z-dn.net/?f=%5Clog_bc%3D-%5Cdfrac34%5Cimplies%20b%5E%7B-3%2F4%7D%3Dc%5Cimplies%20b%3Dc%5E%7B-4%2F3%7D)
![\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12](https://tex.z-dn.net/?f=%5Clog_cd%3D2%5Cimplies%20c%5E2%3Dd%5Cimplies%5Clog_dc%5E2%3D1%5Cimplies%5Clog_dc%3D%5Cdfrac12)
Then
![abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}](https://tex.z-dn.net/?f=abc%3D%28c%5E%7B-4%2F3%7D%29%5E%7B9%2F8%7Dc%5E%7B-4%2F3%7Dc%3Dc%5E%7B-11%2F6%7D)
So we have
![\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}](https://tex.z-dn.net/?f=%5Clog_dabc%3D%5Clog_dc%5E%7B-11%2F6%7D%3D-%5Cdfrac%7B11%7D6%5Clog_dc%3D-%5Cdfrac%7B11%7D6%5Ccdot%5Cdfrac12%3D-%5Cdfrac%7B11%7D%7B12%7D)
Answer:
Step-by-step explanation:
In statistical hypothesis testing we use p values. For any test we select significance level alpha as 5% or 1% or 10%
Whenever p < alpha we reject our null hypothesis. Otherwise we accept our null hypothesis.
Hence p should be high to accept null hypothesis
Option A is incorrect
B is correct
C is wrong as p is the probability
D False
Option B is right.
Do cross multiplication.
Convert ounce to pound.
8 ounce is 0.5 lbs
x/0.5 = 17.95/1
x=8.975
Miss Lawrence spent about $8.98
To be parallel with that line the equation has to have the same slope as the equation given, like y = 10x + 3.